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# expression and evaluation order and precedence rules

hi,

i am a bit confused on expression evaluation order in expressions
involving unary increment.decrement operators along with binary
operators.

For example in the following expression

x += i + j + k++;

is the following the sequence of evaluations

1) as ++ has highest precedencce subexpression k++ gets evaluated
2) i + j is evaluated next and say the result is stored in a temporary
location say temp
3) next temp + k++ gets evaluated and result stored in temporary
location say temp1
4) next x + temp1 gets evaluated and the result is assigned back to x.

So is the sequence of evaluations correct or am i missing out
something here?
Would it be correct to assume tht given an expression all unary
oerators are evaluated followed by evaluation of binary operators
thanks,
silpa

Jun 2 '07 #1
32 2927
si****@gmail.com said:
hi,

i am a bit confused on expression evaluation order in expressions
involving unary increment.decrement operators along with binary
operators.

For example in the following expression

x += i + j + k++;

is the following the sequence of evaluations

1) as ++ has highest precedencce subexpression k++ gets evaluated
No. You can think of precedence and associativity as sorting out which
expressions get to be the operands for any given operator. Precedence
and associativity rules mean that the above expression has the same
meaning as:

x += ((i + j) + (k++));

The order of evaluation of the expressions is unspecified. k++ might be
done first, or it might not. C doesn't say. What it does say, however,
is that the result of k++ is the value k had at the previous sequence
point, so it turns out that it doesn't actually matter anyway.

I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.

Would it be correct to assume tht given an expression all unary
oerators are evaluated followed by evaluation of binary operators
No. See above.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Jun 2 '07 #2
Richard Heathfield wrote:
I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.
(fx:nitpick) False, since the P & A determines what's an operand of
what, and in general operands have to be evaluated before the operation
of which they are a part.

But that's only an indirect effect.

--
"There's a doorway where there was a wall" /Master Humpries Clock/

Hewlett-Packard Limited registered no:
registered office: Cain Road, Bracknell, Berks RG12 1HN 690597 England

Jun 4 '07 #3
Chris Dollin said:
Richard Heathfield wrote:
>I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.

(fx:nitpick) False,
Wrong.
since the P & A determines what's an operand of what,
True.
and in general operands have to be evaluated before the
operation of which they are a part.
That doesn't matter for my point. For example, consider the expression:

a = b() * c() + d();

Here is one legal possibility for the generated assembly language code:

JSR _d
MOV R20, R10 ; copy result of d() to reg 20
JSR _c
MOV R21, R10 ; copy result of c() to reg 21
JSR _b
MOV R22, R10 ; copy result of b() to reg 22
MOV R24, 0 ; reg 24 will hold the result of the entire expr
MUL R21, R22
ADD R24, R21 ; the calculation is now complete
MOV _a, R24

Here, the order of evaluation is d(), c(), b(), addition,
multiplication, whereas the precedence and associativity ordering would
suggest pretty much the opposite.
But that's only an indirect effect.

I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Jun 4 '07 #4
In article <7Z******************************@bt.com>,
Richard Heathfield <rj*@see.sig.invalidwrote:
>Chris Dollin said:
>and in general operands have to be evaluated before the
operation of which they are a part.
>That doesn't matter for my point. For example, consider the expression:

a = b() * c() + d();
[...]
>I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.
I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and that
was true even in your example code.

Of course, you might be able to generate code sufficiently different
from the original expression that nothing can be identified as "the
addition corresponding to the '+' operator".

-- Richard

--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
Jun 4 '07 #5
Richard Tobin said:
In article <7Z******************************@bt.com>,
Richard Heathfield <rj*@see.sig.invalidwrote:
>For example, consider the expression:

a = b() * c() + d();
[...]
>>I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.

I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and that
was true even in your example code.
My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and associativity
rules would suggest it should actually be evaluated *last* (if they
were even remotely relevant).
Of course, you might be able to generate code sufficiently different
from the original expression that nothing can be identified as "the
addition corresponding to the '+' operator".
Indeed. In fact, it would be easy - just replace the addition with an
AND, an XOR, a LSH, and a jump, with a Russian-peasant-multiplication
loop round the outside. Then you can mix all the additions and
multiplications up into one sorry mess in which it would be
unreasonable to say "this happens, and then that happens", because they
overlap so much.

I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Jun 4 '07 #6
Richard Heathfield wrote:
Chris Dollin said:
>Richard Heathfield wrote:
>>I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.

(fx:nitpick) False,

Wrong.
>since the P & A determines what's an operand of what,

True.
>and in general operands have to be evaluated before the
operation of which they are a part.

That doesn't matter for my point. For example, consider the expression:

a = b() * c() + d();

Here is one legal possibility for the generated assembly language code:
(fx:snip)
Here, the order of evaluation is d(), c(), b(), addition,
multiplication, whereas the precedence and associativity ordering would
suggest pretty much the opposite.
(fx:nitpick) False.

Given the expression b() * c() + d(), it is /exactly/ the precedence
that determines that the addition is done after the multiplication
(in the C abstract machine) -- because one operand of the addition
is the result of the multiplication.

It doesn't determine the order of those particular /operands/, of course;
that doesn't matter, because of "nitpick" and-jointly "*anything*".

(I share with you the idea that it's important for newbies to learn
that P&A doesn't determine OOE; but your extreme wording leaves your
lesson more subject to attack than a less extreme wording would. IMAO.)

--
"You're not supposed to /think/ about it, /The Beiderbeck Connection/
you're supposed to say NO!" Jill Swinburn

Hewlett-Packard Limited registered no:
registered office: Cain Road, Bracknell, Berks RG12 1HN 690597 England

Jun 4 '07 #7
In article <2M******************************@bt.com>,
Richard Heathfield <rj*@see.sig.invalidwrote:
>>>a = b() * c() + d();
>>>I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.
>I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and that
was true even in your example code.
>My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and associativity
rules would suggest it should actually be evaluated *last* (if they
were even remotely relevant).
That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication. At
best you can say there might not be any order, but some orders are
not possible.

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
Jun 4 '07 #8
Chris Dollin said:
Richard Heathfield wrote:
>Chris Dollin said:
>>Richard Heathfield wrote:

I cannot stress this enough: neither precedence nor associativity
has *anything* to do with order of evaluation.

(fx:nitpick) False,

Wrong.
>>since the P & A determines what's an operand of what,

True.
>>and in general operands have to be evaluated before the
operation of which they are a part.

That doesn't matter for my point. For example, consider the
expression:

a = b() * c() + d();

Here is one legal possibility for the generated assembly language
code:

(fx:snip)
>Here, the order of evaluation is d(), c(), b(), addition,
multiplication, whereas the precedence and associativity ordering
would suggest pretty much the opposite.

(fx:nitpick) False.

Given the expression b() * c() + d(), it is /exactly/ the precedence
that determines that the addition is done after the multiplication
(in the C abstract machine) -- because one operand of the addition
is the result of the multiplication.
Neither the C abstract machine nor the Standard impose an order of
evaluation on the implementation. For example, 2 * 2 + 2 need not
result in any additions or multiplications being carried out. The
implementation could easily code this as (2 << 1) | 2 if it wanted. Or
even as 6, with no runtime evaluation happening at all!
It doesn't determine the order of those particular /operands/, of
course; that doesn't matter, because of "nitpick" and-jointly
"*anything*".
Right - it doesn't determine the order of those operands. Neither does
it determine the *order in which they are evaluated*.
(I share with you the idea that it's important for newbies to learn
that P&A doesn't determine OOE; but your extreme wording leaves your
lesson more subject to attack than a less extreme wording would.
IMAO.)
Oh, attack away, my dear chap - with any luck, you'll prove I'm wrong,
and then I'll have learned something.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Jun 4 '07 #9
Richard Tobin said:
In article <2M******************************@bt.com>,
Richard Heathfield <rj*@see.sig.invalidwrote:
>>>>a = b() * c() + d();
>>>>I cannot stress this enough: neither precedence nor associativity
has *anything* to do with order of evaluation.
>>I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and that
was true even in your example code.
>>My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and associativity
rules would suggest it should actually be evaluated *last* (if they
were even remotely relevant).

That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication.
Okay, I agree that I've overstated my case (or at least can't prove I
haven't). My apologies to you and Chris.
At
best you can say there might not be any order,
That is the point I was trying to make.
but some orders are not possible.
I spent a good five minutes trying to construct a counter-example, and
failed. This may be due to my lack of imagination, of course, but the
more likely explanation is that you are correct.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Jun 4 '07 #10
In article <2M******************************@bt.com>, Richard Heathfield
<rj*@see.sig.invalidwrites
>
I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.
You are correct.
Precedence is laid down in the standard
Associatively is also defined in the standard

The Order of Evaluation is NOT defined in the standard.
A line of C code can be evaluated left to right, right to left, middle
out or any order.

This has caused problems in the past. The most common single method is
Left to right. I am not sure if it has a majority though.

--
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
\/\/\/\/\ Chris Hills Staffs England /\/\/\/\/
/\/\/ ch***@phaedsys.org www.phaedsys.org \/\/\
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/

Jun 4 '07 #11
Richard Tobin wrote:
In article <2M******************************@bt.com>,
Richard Heathfield <rj*@see.sig.invalidwrote:
>>>a = b() * c() + d();
>>>I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.
>>I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and that
was true even in your example code.
>My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and associativity
rules would suggest it should actually be evaluated *last* (if they
were even remotely relevant).

That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication. At
best you can say there might not be any order, but some orders are
not possible.
Precede the above with

#define d() 0

.... or with any definition of d() that allows the optimizer
to discover that no side-effects are produced and the value
returned is zero. Now, with as straight a face as you can
muster, tell us all again that the addition must come after
the multiplication. Chances are, the addition will occur at
compile time and thus precede the multiplication by a *lot*.

#define d() c()

.... and suppose a c() that the optimizer can see is "pure."
There's a decent chance that the compiler may generate the
same code as for

a = (b() + 1) * c();

Muster that straight face yet again, please.

--
Eric Sosman
es*****@acm-dot-org.invalid
Jun 4 '07 #12
In article <M9*********************@bt.com>, Richard Heathfield
<rj*@see.sig.invalidwrites
>Richard Tobin said:
>In article <2M******************************@bt.com>,
Richard Heathfield <rj*@see.sig.invalidwrote:
>>>>>a = b() * c() + d();
>>>>>I cannot stress this enough: neither precedence nor associativity
>has *anything* to do with order of evaluation.
>>>I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and that
was true even in your example code.
>>>My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and associativity
rules would suggest it should actually be evaluated *last* (if they
were even remotely relevant).

That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication.

Okay, I agree that I've overstated my case (or at least can't prove I
haven't). My apologies to you and Chris.
>At
best you can say there might not be any order,

That is the point I was trying to make.
>but some orders are not possible.

I spent a good five minutes trying to construct a counter-example, and
failed. This may be due to my lack of imagination, of course, but the
more likely explanation is that you are correct.
This one is interesting but a bit of a red herring...
http://www.phaedsys.demon.co.uk/chri...swengtips3.htm
and
http://www.phaedsys.demon.co.uk/chri...wengtips3a.htm

It clearly shows that you have to take care. :-)

--
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
\/\/\/\/\ Chris Hills Staffs England /\/\/\/\/
/\/\/ ch***@phaedsys.org www.phaedsys.org \/\/\
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/

Jun 4 '07 #13
Eric Sosman wrote:
Richard Tobin wrote:
>That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication. At
best you can say there might not be any order, but some orders are
not possible.

Precede the above with

#define d() 0

... or with any definition of d() that allows the optimizer
to discover that no side-effects are produced and the value
returned is zero. Now, with as straight a face as you can
muster, tell us all again that the addition must come after
the multiplication. Chances are, the addition will occur at
compile time and thus precede the multiplication by a *lot*.
Sure. That's invisible to the abstract machine, though, as
are all the as-ifs.

(I'm interpreting Richard T's "can't" as "can't, in general",
which is what I'd mean if I'd written it in this discussion.)

--
"There's a doorway where there was a wall" /Master Humpries Clock/

Hewlett-Packard Limited registered no:
registered office: Cain Road, Bracknell, Berks RG12 1HN 690597 England

Jun 4 '07 #14
Chris Dollin wrote:
>
Richard Heathfield wrote:
I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.

(fx:nitpick) False, since the P & A determines what's an operand of
what, and in general operands have to be evaluated before the operation
of which they are a part.

But that's only an indirect effect.
Yes, in order to evaluate "a + b", the values of "a" and "b" must be
determined before the addition takes place. But, there is nothing
that says anything about which gets evaluated first. The order of
evaluation is not specified.

Taking this further:

( a + b ) * ( c - d )

Again, "a" and "b" must be evaluated some time before the addition
is done, and "c" and "d" must be evaluated some time before the
subtraction is done, and both the addition and subtraction must be[*]
done some time before the multiplication is done. However, there
is nothing which specifies the order in which a, b, c, and d are
evaluated.
[*] In fact, as far as I understand it, nothing really requires
that "a+b" or "c-d" be evaluated before the multiplication is done,
as long as the result is the same. Although inefficient, I believe
it is possible for the compiler to treat the above as if it were
written as:

a*c + b*c - a*d - b*d;

as long as any side effects aren't duplicated.

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody | www.hvcomputer.com | #include |
| kenbrody/at\spamcop.net | www.fptech.com | <std_disclaimer.h|
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <mailto:Th*************@gmail.com>
Jun 4 '07 #15
In article <G5******************************@comcast.com>,
Eric Sosman <es*****@acm-dot-org.invalidwrote:
>That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication. At
best you can say there might not be any order, but some orders are
not possible.
Precede the above with

#define d() 0
[...]
Well obviously. But all that means is that there are some more cases
where the order isn't fixed, when the claim I was addressing is that
precedence *never* constrains evaluation order.

Richard Heathfield is entirely right to separate precedence and
associativity - which apply to the mapping from program syntax to
semantics - from evaluation order, which can be chosen in any way that
implements the semantics. He just went too far in denying that the
syntactic features have *any* consequences for the evaluation order.

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
Jun 4 '07 #16
In article <LF**************@phaedsys.demon.co.uk>,
Chris Hills <ch***@phaedsys.demon.co.ukwrote:
>The Order of Evaluation is NOT defined in the standard.
A line of C code can be evaluated left to right, right to left, middle
out or any order.

This has caused problems in the past. The most common single method is
Left to right. I am not sure if it has a majority though.
For most compilers, within an expression, optimisation will choose
factors such as making best use of registers, so that choice of
syntactic order for equivalent expressions will have little if any
effect.

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.
Jun 4 '07 #17
Kenneth Brody wrote:
Chris Dollin wrote:
>>
Richard Heathfield wrote:
I cannot stress this enough: neither precedence nor associativity has
*anything* to do with order of evaluation.

(fx:nitpick) False, since the P & A determines what's an operand of
what, and in general operands have to be evaluated before the operation
of which they are a part.

But that's only an indirect effect.

Yes, in order to evaluate "a + b", the values of "a" and "b" must be
determined before the addition takes place. But, there is nothing
that says anything about which gets evaluated first. The order of
evaluation is not specified.
The order of evaluation /of the operands/ is not specified; I haven't
claimed otherwise.

In a + b, there's no room for P & A to make a difference.

In a + b * c, without P & A [1] there are two possible unambiguous
readings: (a + b) * c, and a + (b * c). In both of these, the
order of the evaluation of the /operands/ a, b, c is unspecified.
However, in one the result of the addition is an operand of the
multiplication, and so /in general/ [2] the addition (not the
evaluation of its operands, the addition itself) has to be
done before the multiplication; and t'other way about for the
other one.

[1] And ignoring the detail that C proper doesn't use P&A at all,
and the grammar gives an unambiguous reading to a + b * c.

[2] ie one can construct /specific/ counter-examples, but without
that specicifity you can't assume otherwise.

--
"Our future looks secure, but it's all out of our hands." /Man and Machine/
- Magenta

Hewlett-Packard Limited Cain Road, Bracknell, registered no:
registered office: Berks RG12 1HN 690597 England

Jun 4 '07 #18
Richard Tobin wrote:
>
For most compilers, within an expression, optimisation will choose
factors such as making best use of registers, so that choice of
syntactic order for equivalent expressions will have little if any
effect.
I have seen cases where the order of evaluation changed between debug
(fast - non optimizing compiler) and release (optimizing compiler). This
was at the time when I was not entirely aware that it could change
myself and therefore introduced an error - of course the error only
existed in the hard to debug release version of the program.
Jun 4 '07 #19
Richard Heathfield <r...@see.sig.invalidwrote:
sil...@gmail.com said:
i am a bit confused on expression evaluation order in
expressions involving unary increment.decrement operators
along with binary operators.

For example in the following expression

x += i + j + k++;

is the following the sequence of evaluations

1) as ++ has highest precedencce subexpression k++ gets
evaluated

No. You can think of precedence and associativity as sorting
out which expressions get to be the operands for any given
operator. Precedence and associativity rules mean that the
above expression has the same meaning as:

x += ((i + j) + (k++));

The order of evaluation of the expressions is unspecified.
The order of sub-expressions is unspecified, except for ||, &&
and ?: operators. As you point out yourself, there is an implied
ordering of the + operations from the associativity.

<snip>
I cannot stress this enough: neither precedence nor
associativity has *anything* to do with order of
evaluation.
The expression i + j + k++ is evaluated on the virtual machine
as (i + j) + k++, not as i + (j + k++). There is potentially
a real difference on some machines.

Given...

#define CONV (-'0')

....and assuming *p is a digit, consider...

n = *p + CONV + n * 10;

This is a correct form only because there is an implied
ordering of evaluation. Without the implied ordering, I
would have to use parentheses, or break the statement
into multiple statements.

[Of course in practice I would write n = n * 10 + (*p - '0'),
but that doesn't alter the above comments.]

--
Peter

Jun 5 '07 #20
Peter Nilsson said:
Richard Heathfield <r...@see.sig.invalidwrote:
>sil...@gmail.com said:
i am a bit confused on expression evaluation order in
expressions involving unary increment.decrement operators
along with binary operators.

For example in the following expression

x += i + j + k++;

is the following the sequence of evaluations

1) as ++ has highest precedencce subexpression k++ gets
evaluated

No. You can think of precedence and associativity as sorting
out which expressions get to be the operands for any given
operator. Precedence and associativity rules mean that the
above expression has the same meaning as:

x += ((i + j) + (k++));

The order of evaluation of the expressions is unspecified.

The order of sub-expressions is unspecified, except for ||, &&
and ?: operators. As you point out yourself, there is an implied
ordering of the + operations from the associativity.
I didn't point out any such thing. The associativity determines which
operands are used with which operators. That isn't the same as implying
an order of evaluation. Any evaluation ordering that happens to fall
out of an expression syntax is purely a consequence of reality, and has
nothing to do with the rules of C.

<snip>
>I cannot stress this enough: neither precedence nor
associativity has *anything* to do with order of
evaluation.

The expression i + j + k++ is evaluated on the virtual machine
as (i + j) + k++, not as i + (j + k++). There is potentially
a real difference on some machines.
If the values are initially, say, these:

i = 1;
j = -1;
k = INT_MIN;

then there is a real difference in the abstract machine, too.
Given...

#define CONV (-'0')

...and assuming *p is a digit, consider...

n = *p + CONV + n * 10;

This is a correct form only because there is an implied
ordering of evaluation.
Why? Precedence looks perfectly adequate here.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Jun 5 '07 #21
Richard Heathfield wrote:
>
.... snip ...
>
Neither the C abstract machine nor the Standard impose an order of
evaluation on the implementation. For example, 2 * 2 + 2 need not
result in any additions or multiplications being carried out. The
implementation could easily code this as (2 << 1) | 2 if it wanted.
Or even as 6, with no runtime evaluation happening at all!
Now that is overstating things. (2 << 1) is defined as 2 * 2, i.e.
as am arithmetic operation.

--
<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>
cbfalconer at maineline dot net

--
Posted via a free Usenet account from http://www.teranews.com

Jun 6 '07 #22
Chris Dollin wrote:
Richard Heathfield wrote:
>I cannot stress this enough: neither precedence nor associativity
has *anything* to do with order of evaluation.

(fx:nitpick) False, since the P & A determines what's an operand
of what, and in general operands have to be evaluated before the
operation of which they are a part.

But that's only an indirect effect.
Oh? For example, with sizeof int == 2, the addition operation
could well be simplified to:

load lsb1; the digit refers to the variable location
store lsbresult; (and save carry)
adc msb2; note using the carry
store msbresult; A carry now implies overflow.

and neither operand was pre-evaluated. This will work just fine on
simple cpus such as the PIC that have a single 1-byte arithmetic
register, the accumulator.

--
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Jun 6 '07 #23
Richard Tobin wrote:
Richard Heathfield <rj*@see.sig.invalidwrote:
*** somebody lost attributions - evil **
>
>>>a = b() * c() + d();
>>>I cannot stress this enough: neither precedence nor associativity
has *anything* to do with order of evaluation.
>>I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and
that was true even in your example code.
>My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and
associativity rules would suggest it should actually be evaluated
*last* (if they were even remotely relevant).

That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication. At
best you can say there might not be any order, but some orders are
not possible.
I think people are missing something. Read my earlier observation
on adding (a few minutes ago) first, and then consider A * B in a
machine with sizeof int == 2. We do need an 8*8 multiply, and
interior carries have to be saved and used.

We can form A * B via:
lsbA * lsbB + K*(lsbA * msbB + lsbB * msbA) + K*K*(msbA * msbB)

and there are better algorithms available. K is related to
CHARBIT. So design the 8*8 multiply (when CHARBIT == 8) and you
have it all.

It helps to have built all this before.

--
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cbfalconer at maineline dot net

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Jun 6 '07 #24
Richard Heathfield <r...@see.sig.invalidwrote:
Peter Nilsson said:
Richard Heathfield <r...@see.sig.invalidwrote:
The order of evaluation of the expressions is
unspecified.
The order of sub-expressions is unspecified, ...
Your statement is too strong. 6.5p3 says sub-expressions,
not expressions, for a reason IMO.
I cannot stress this enough: neither precedence nor
associativity has *anything* to do with order of
evaluation.
The expression i + j + k++ is evaluated on the virtual
machine as (i + j) + k++, not as i + (j + k++). There
is potentially a real difference on some machines.

If the values are initially, say, these:

i = 1;
j = -1;
k = INT_MIN;

then there is a real difference in the abstract machine,
too.
I cannot understand your statement that the ordering of
evaluation does not have _anything_ to do with associativity
when you are acknowledging the very fact that there is an
implied ordering on the abstract machine based on
associativity.

Perhaps I'm deluding myself, but I put it to you that my
(shared) delusion has practical usefulness that your
literism denies.

I write...

n = n * 10 + (*p - '0');

....because I believe the additive operators in...

n = n * 10 + *p - '0';

....are evaluated left to right and render the latter
expression flawed.

If I _had_ to reject the concept of associativity giving
an order to evaluation in that expression, I doubt I'd
continue programming.

Similarly, I don't think of expressions in terms of C
grammar, I think of them in terms of precedence and
associativity tables. I do so because the latter is
equivalent but much easier to apply in practice.

--
Peter

Jun 6 '07 #25
CBFalconer wrote:
Richard Tobin wrote:
>Richard Heathfield <rj*@see.sig.invalidwrote:
*** somebody lost attributions - evil **
>>
>>>>a = b() * c() + d();
>>>>I cannot stress this enough: neither precedence nor associativity
has *anything* to do with order of evaluation.
>>>I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and
that was true even in your example code.
>>My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and
associativity rules would suggest it should actually be evaluated
*last* (if they were even remotely relevant).

That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication. At
best you can say there might not be any order, but some orders are
not possible.

I think people are missing something.
Yes; I have no idea how your post relates to the context; you'll
have to unpack the relationship for me.

--
"We live for the One, you die for the One." Unsaid /Babylon 5/

Hewlett-Packard Limited registered office: Cain Road, Bracknell,
registered no: 690597 England Berks RG12 1HN

Jun 6 '07 #26
Chris Dollin wrote:
CBFalconer wrote:
>Richard Tobin wrote:
>>Richard Heathfield <rj*@see.sig.invalidwrote:
*** somebody lost attributions - evil **
>>>
>a = b() * c() + d();

>I cannot stress this enough: neither precedence nor associativity
>has *anything* to do with order of evaluation.

I think Chris's point is that the relative precedence of + and *
forces the multiplication corresponding to the "*" operators to be
done before the addition corresponding to the "+" operator, and
that was true even in your example code.

My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and
associativity rules would suggest it should actually be evaluated
*last* (if they were even remotely relevant).

That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication. At
best you can say there might not be any order, but some orders are
not possible.

I think people are missing something.

Yes; I have no idea how your post relates to the context; you'll
have to unpack the relationship for me.
I don't remember the details of the post, and you neglected to
quote it. So I have no reply.

--
<http://www.cs.auckland.ac.nz/~pgut001/pubs/vista_cost.txt>
<http://www.securityfocus.com/columnists/423>
<http://www.aaxnet.com/editor/edit043.html>
cbfalconer at maineline dot net

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Jun 6 '07 #27
CBFalconer wrote:
Chris Dollin wrote:
>CBFalconer wrote:
>>Richard Tobin wrote:
Richard Heathfield <rj*@see.sig.invalidwrote:

*** somebody lost attributions - evil **

>>a = b() * c() + d();

>>I cannot stress this enough: neither precedence nor associativity
>>has *anything* to do with order of evaluation.

>I think Chris's point is that the relative precedence of + and *
>forces the multiplication corresponding to the "*" operators to be
>done before the addition corresponding to the "+" operator, and
>that was true even in your example code.

My point is that one of the operands of the addition was actually
evaluated *first*, despite the fact that precedence and
associativity rules would suggest it should actually be evaluated
*last* (if they were even remotely relevant).

That's not enough to show that precedence has *nothing* to do with
evaluation order. b(), c(), and d() can be evaluated in any order,
but the addition can't be evaluated before the multiplication. At
best you can say there might not be any order, but some orders are
not possible.

I think people are missing something.

Yes; I have no idea how your post relates to the context; you'll
have to unpack the relationship for me.

I don't remember the details of the post, and you neglected to
quote it. So I have no reply.
\begin{quote}
I think people are missing something. Read my earlier observation
on adding (a few minutes ago) first, and then consider A * B in a
machine with sizeof int == 2. We do need an 8*8 multiply, and
interior carries have to be saved and used.

We can form A * B via:
lsbA * lsbB + K*(lsbA * msbB + lsbB * msbA) + K*K*(msbA * msbB)

and there are better algorithms available. K is related to
CHARBIT. So design the 8*8 multiply (when CHARBIT == 8) and you
have it all.

It helps to have built all this before.
\end{quote}

I don't see how this relates to precedence vs order-of-evaluation.

--
"Reaching out for mirrors hidden in the web." - Renaissance, /Running Hard/

Hewlett-Packard Limited registered no:
registered office: Cain Road, Bracknell, Berks RG12 1HN 690597 England

Jun 7 '07 #28
Chris Dollin wrote:
>
CBFalconer wrote:
.... snip ...
>
>I don't remember the details of the post, and you neglected to
quote it. So I have no reply.

\begin{quote}
I think people are missing something. Read my earlier observation
on adding (a few minutes ago) first, and then consider A * B in a
machine with sizeof int == 2. We do need an 8*8 multiply, and
interior carries have to be saved and used.

We can form A * B via:
lsbA * lsbB + K*(lsbA * msbB + lsbB * msbA) + K*K*(msbA * msbB)

and there are better algorithms available. K is related to
CHARBIT. So design the 8*8 multiply (when CHARBIT == 8) and you
have it all.

It helps to have built all this before.
\end{quote}

I don't see how this relates to precedence vs order-of-evaluation.
Nor do I :-)

--
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<http://www.aaxnet.com/editor/edit043.html>
cbfalconer at maineline dot net

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Jun 7 '07 #29
On 4 Jun 2007 at 22:42, CBFalconer wrote:
Oh? For example, with sizeof int == 2, the addition operation
could well be simplified to:

load lsb1; the digit refers to the variable location
store lsbresult; (and save carry)
adc msb2; note using the carry
store msbresult; A carry now implies overflow.

and neither operand was pre-evaluated. This will work just fine on
simple cpus such as the PIC that have a single 1-byte arithmetic
register, the accumulator.
Assembly is off-topic in c.l.c. So is pseudo-assembly.
How (if at all) the implementation chooses to translate code to
machine-language is inherently non-portable and non-standard.
--
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<http://www.aaxnet.com/editor/edit043.html>
cbfalconer at maineline dot net

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Your signature is in gross violation of Usenet conventions and basic

Jun 14 '07 #30
In article <sl*********************@nospam.com>, Net Nanny
<ne******@pointlesspedantry.comwrites
>Your signature is in gross violation of Usenet conventions and basic
Your post was also a gross violation of Usenet conventions and basic
Netiquette

You must use ONLY ASCII for this NG

--
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\
\/\/\/\/\ Chris Hills Staffs England /\/\/\/\/
/\/\/ ch***@phaedsys.org www.phaedsys.org \/\/\
\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/

Jun 14 '07 #31
Net Nanny wrote:
>
On 4 Jun 2007 at 22:42, CBFalconer wrote:
Oh? For example, with sizeof int == 2, the addition operation
could well be simplified to:
[... snip pseudo-assembly example ...]

and neither operand was pre-evaluated. This will work just fine on
simple cpus such as the PIC that have a single 1-byte arithmetic
register, the accumulator.

Assembly is off-topic in c.l.c. So is pseudo-assembly.
How (if at all) the implementation chooses to translate code to
machine-language is inherently non-portable and non-standard.
I don't believe it's OT when being used as an example of a valid
compilation output, especially when the discussion was about what
sort of results are valid. For example, some of the myriad
possible outputs for "foo(i++,i++)", or how "i=i++" could result
in a frozen computer.

Sometimes, the best way to describe a situation is to give some
[pseudo-]assembler code.

[...]

--
+-------------------------+--------------------+-----------------------+
| Kenneth J. Brody | www.hvcomputer.com | #include |
| kenbrody/at\spamcop.net | www.fptech.com | <std_disclaimer.h|
+-------------------------+--------------------+-----------------------+
Don't e-mail me at: <mailto:Th*************@gmail.com>

Jun 15 '07 #32
On Fri, 15 Jun 2007 10:24:10 -0400, in comp.lang.c , Kenneth Brody
<ke******@spamcop.netwrote:
>Net Nanny wrote:
>>
Assembly is off-topic in c.l.c. So is pseudo-assembly.
How (if at all) the implementation chooses to translate code to
machine-language is inherently non-portable and non-standard.

I don't believe it's OT when being used as an example of a valid
compilation output,
Don't waste your breath - NetNanny is a troll, probably a sockpuppet
for one of our resident ones.
--
Mark McIntyre

"Debugging is twice as hard as writing the code in the first place.
Therefore, if you write the code as cleverly as possible, you are,
by definition, not smart enough to debug it."
--Brian Kernighan
Jun 15 '07 #33

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