About sizeof

[³á³á³á³á]

int i = 0;
i = sizeof( i++ );

What's the result of i¡H

I think the result may be sizeof( int ) or sizeof( int ) + 1, or it is

undefined behavior.

I know that i = i++ is undefined behavior,

but sizeof is calculated at compiling time.

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¡þ plum.cs.nccu.edu.tw

[Tom Zych]

Someone whose name my newsreader would probably mangle wrote:

int i = 0;
i = sizeof( i++ ); I think the result may be sizeof( int ) or sizeof( int ) + 1, or it is
undefined behavior. I know that i = i++ is undefined behavior,
but sizeof is calculated at compiling time.

But you're still changing the value of i twice in one statement[1].
i++ increments it. i = sizeof... sets it to a constant value.
Undefined behavior. Anything can happen. The computer can whistle
"Night in Tunisia", make a pot of coffee, and explode.

This ignores the question of whether sizeof (i++) is legal or
well-defined. I don't know that one.

[1] To be precise, without an intervening sequence point.

--
Tom Zych
This email address will expire at some point to thwart spammers.
Permanent address: echo 'g******@cbobk.pbz' | rot13

[Jirka Klaue]

Tom Zych wrote:

Someone whose name my newsreader would probably mangle wrote:

int i = 0;
i = sizeof( i++ );

I think the result may be sizeof( int ) or sizeof( int ) + 1, or it is
undefined behavior.

I know that i = i++ is undefined behavior,
but sizeof is calculated at compiling time.

But you're still changing the value of i twice in one statement[1].
i++ increments it. i = sizeof... sets it to a constant value.
Undefined behavior. Anything can happen. The computer can whistle
"Night in Tunisia", make a pot of coffee, and explode.

No, sizeof does not evaluate i++.

6.5.3.4 The sizeof operator
Semantics
2 The sizeof operator yields the size (in bytes) of its operand, which may be an
expression or the parenthesized name of a type. The size is determined from the
type ofthe operand. The result is an integer. If the type of the operand is a
variable length array type, the operand is evaluated; otherwise, the operand is
not evaluated and the result is an integer constant.
This ignores the question of whether sizeof (i++) is legal or
well-defined. I don't know that one.

The type of i++ is int, therefore it's equivalent to sizeof(int).

Jirka

[Johan Aurér]

On Thu, 11 Sep 2003, Tom Zych wrote:

Someone whose name my newsreader would probably mangle wrote:

int i = 0;
i = sizeof( i++ );

I think the result may be sizeof( int ) or sizeof( int ) + 1, or it is
undefined behavior.

I know that i = i++ is undefined behavior,
but sizeof is calculated at compiling time.

But you're still changing the value of i twice in one statement[1].
i++ increments it. i = sizeof... sets it to a constant value.

No, sizeof does not evaluate its operand, so sizeof( i++ ) is equivalent
to sizeof( i ).

--
au***@axis.com

[Dan Pop]

In <48********@shepjeng.twbbs.org> GA************@shepjeng.twbbs.org (³á³á³á³á) writes:

int i = 0;
i = sizeof( i++ );

What's the result of i¡H

I think the result may be sizeof( int ) or sizeof( int ) + 1, or it is
undefined behavior.
It's sizeof(int). When the argument of sizeof is an expression (other
than a C99 VLA), it is NOT evaluated, the compiler merely determines its
type and produces the appropriate result.
I know that i = i++ is undefined behavior,
This is a completely different issue.
but sizeof is calculated at compiling time.

It need not be, unless used in a context where a constant expression is
required.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Jirka Klaue]

Johan Aurér wrote:
....

No, sizeof does not evaluate its operand, so sizeof( i++ ) is equivalent
to sizeof( i ).

That's true in this case, but not generally:

short i;
assert( sizeof(i) == sizeof (i + 1) );

Jirka

[Jeff]

"³á³á³á³á" <GA************@shepjeng.twbbs.org> wrote in message
news:48********@shepjeng.twbbs.org...

int i = 0;
i = sizeof( i++ );

What's the result of i¡H

Using sizeof to "i++ " is similar to "sizeof(int)". i++ returns an integer,
so your compiler just think about "what is the size of integer in this
implementation"

try this

printf("The size is %d \n", sizeof(i*32/9 + 2525 - i++ * i--));
--
Jeff
-je6543 at yahoo.com

[Tom Zych]

Jirka Klaue wrote:

No, sizeof does not evaluate i++.

Oh. Ok, my bad.

Have to remember that sizeof(i++) for the IOCCC. It's no good for
anything else :)

--
Tom Zych
This email address will expire at some point to thwart spammers.
Permanent address: echo 'g******@cbobk.pbz' | rot13

[Derk Gwen]

# int i = 0;
# i = sizeof( i++ );
#
# What's the result of i¡H

'Doctor! Doctor! It hurts when I do this!'
'Then don't that.'

Is this like a review of obfustucated C contest code, or exams of a sadistic
instructor, or malevolent job interviewers? I see people bring up truly twisted
code that no programmer would ever use in production code, and then ask
'what does this do?' I'm left wonderring, why do this people even want to know?

--
Derk Gwen http://derkgwen.250free.com/html/index.html
But I do believe in this.

[Jack Klein]

On Thu, 11 Sep 2003 19:14:34 +0200, Jirka Klaue
<jk****@ee.tu-berlin.de> wrote in comp.lang.c:

Johan Aurér wrote:
...

No, sizeof does not evaluate its operand, so sizeof( i++ ) is equivalent
to sizeof( i ).

That's true in this case, but not generally:

short i;
assert( sizeof(i) == sizeof (i + 1) );

Jirka

Works just fine on the TI DSP compiler I'm working with at the moment.

If fact this one will not assert either, on this particular
implementation:

char i;
assert (sizeof(i) == sizeof(i + 1));

CHAR_BIT is 16 and char, short, and int all have identical
representations.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++ ftp://snurse-l.org/pub/acllc-c++/faq

[Mantorok Redgormor]

Jirka Klaue <jk****@ee.tu-berlin.de> wrote in message news:<3F**************@ee.tu-berlin.de>...

Johan Aurér wrote:
...

No, sizeof does not evaluate its operand, so sizeof( i++ ) is equivalent
to sizeof( i ).

That's true in this case, but not generally:

short i;
assert( sizeof(i) == sizeof (i + 1) );

Jirka

Why does sizeof i + 1 give 3 and not the size of an int on my system?
I thought it does not evaluate its operand? This makes no sense.

[Tom Zych]

Mantorok Redgormor wrote:

Why does sizeof i + 1 give 3 and not the size of an int on my system?
I thought it does not evaluate its operand? This makes no sense.

Because sizeof has a higher precedence than +. Your expression is
equivalent to (sizeof i) + 1.

--
Tom Zych
This email address will expire at some point to thwart spammers.
Permanent address: echo 'g******@cbobk.pbz' | rot13

[Irrwahn Grausewitz]

ne*****@tokyo.com (Mantorok Redgormor) wrote:

Jirka Klaue <jk****@ee.tu-berlin.de> wrote in message news:<3F**************@ee.tu-berlin.de>...

Johan Aurér wrote:
...

> No, sizeof does not evaluate its operand, so sizeof( i++ ) is equivalent
> to sizeof( i ).

That's true in this case, but not generally:

short i;
assert( sizeof(i) == sizeof (i + 1) );

Jirka

Why does sizeof i + 1 give 3 and not the size of an int on my system?
I thought it does not evaluate its operand? This makes no sense.

Hehehe, you calculated: ( sizeof i ) + 1 , which gives a result
quite different from: sizeof ( i + 1 ).

Caught by operator precedence... :-)

Irrwahn

--
I can't see it from here, but it looks good to me.

[Jeff]

"Derk Gwen" <de******@HotPOP.com> wrote in message
news:vm************@corp.supernews.com...

# int i = 0;
# i = sizeof( i++ );
#
# What's the result of i¡H

'Doctor! Doctor! It hurts when I do this!'
'Then don't that.'

Is this like a review of obfustucated C contest code, or exams of a sadistic instructor, or malevolent job interviewers? I see people bring up truly twisted code that no programmer would ever use in production code, and then ask
'what does this do?' I'm left wonderring, why do this people even want to know?

When one day someone write such kind of code, and you have to understand it
or explain it to your children :-)

--
Jeff
-je6543 at yahoo.com

[Jirka Klaue]

Jack Klein wrote:

Jirka Klaue wrote:

Johan Aurér wrote:

No, sizeof does not evaluate its operand, so sizeof( i++ ) is equivalent
to sizeof( i ).

That's true in this case, but not generally:

short i;
assert( sizeof(i) == sizeof (i + 1) );

Works just fine on the TI DSP compiler I'm working with at the moment.

If fact this one will not assert either, on this particular
implementation:

char i;
assert (sizeof(i) == sizeof(i + 1));

CHAR_BIT is 16 and char, short, and int all have identical
representations.

That's nice. This compiler isn't a free-standing implementation, is it?
Otherwise EOF is going to be a serious problem.
And I wouldn't call it "general".

Jirka

[Jirka Klaue]

Jirka Klaue wrote:

Jack Klein wrote: ....

char i;
assert (sizeof(i) == sizeof(i + 1));

CHAR_BIT is 16 and char, short, and int all have identical
representations.

That's nice. This compiler isn't a free-standing implementation, is it?

s/free-standing/hosted/
Otherwise EOF is going to be a serious problem.
And I wouldn't call it "general".

Jirka

[Dan Pop]

In <bj**********@mamenchi.zrz.TU-Berlin.DE> Jirka Klaue <jk****@ee.tu-berlin.de> writes:

Jack Klein wrote:

Works just fine on the TI DSP compiler I'm working with at the moment.

If fact this one will not assert either, on this particular
implementation:

char i;
assert (sizeof(i) == sizeof(i + 1));

CHAR_BIT is 16 and char, short, and int all have identical
representations.
That's nice. This compiler isn't a free-standing implementation, is it?

Of course it is. I doubt anyone would attempt a hosted implementation
with UCHAR_MAX > INT_MAX and/or a general purpose computer with a DSP
as its main CPU.
Otherwise EOF is going to be a serious problem.

Because the standard C library specification is written with the implicit
assumption that UCHAR_MAX <= INT_MAX. I've never understood why they
didn't make it explicit.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Alex]

Mantorok Redgormor <ne*****@tokyo.com> wrote:

Jirka Klaue <jk****@ee.tu-berlin.de> wrote in message news:<3F**************@ee.tu-berlin.de>...

Johan Aurer wrote:
...

> No, sizeof does not evaluate its operand, so sizeof( i++ ) is equivalent
> to sizeof( i ).
That's true in this case, but not generally:

short i;
assert( sizeof(i) == sizeof (i + 1) );

Jirka

Why does sizeof i + 1 give 3 and not the size of an int on my system?
I thought it does not evaluate its operand? This makes no sense.

This makes absolute sense.

sizeof i + 1 == (sizeof i) + 1

....which is quite different from:

sizeof (i + 1)

Alex