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associativity of operators

P: n/a
What does "associativity of operators" mean ?

I am unable to undersatand this from K & R 2nd edition. Kindly explain
with an example.

Thanks

Mar 16 '07 #1
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P: n/a
su**************@yahoo.com, India said:
What does "associativity of operators" mean ?

I am unable to undersatand this from K & R 2nd edition. Kindly explain
with an example.

Consider the following code fragment:

int x = 4;
int y = 5;
int z = 6;
int r = x - y - z;

printf("r is %d\n", r);

What would you expect to be printed?

Is it:

(a) -7
(b) 5
(c) something else

If you answered (a), try to work out why (b) is also plausible.
If you answered (b), try to work out why (a) is also plausible.
If you answered (c), come back when you're feeling better. :-)

Which answer you get depends on the order in which the subtractions are
done. 4 - (5 - 6) is different to (4 - 5) - 6. Clearly, precedence
cannot answer this question, since subtraction obviously has the same
precedence as subtraction!

So we need another tool for discriminating between operators at the same
precedence level. That tool is 'associativity'. Left-to-right
associativity means "do the stuff on the left first, and use the
calculated result as input for the stuff on the right", and
right-to-left associativity means, of course, the opposite.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
Mar 16 '07 #2

P: n/a
On Fri, 16 Mar 2007 07:29:43 +0000, Richard Heathfield
<rj*@see.sig.invalidwrote:
>su**************@yahoo.com, India said:
>What does "associativity of operators" mean ?

I am unable to undersatand this from K & R 2nd edition. Kindly explain
with an example.


Consider the following code fragment:

int x = 4;
int y = 5;
int z = 6;
int r = x - y - z;

printf("r is %d\n", r);

What would you expect to be printed?

Is it:

(a) -7
(b) 5
(c) something else

If you answered (a), try to work out why (b) is also plausible.
If you answered (b), try to work out why (a) is also plausible.
If you answered (c), come back when you're feeling better. :-)

Which answer you get depends on the order in which the subtractions are
done. 4 - (5 - 6) is different to (4 - 5) - 6. Clearly, precedence
cannot answer this question, since subtraction obviously has the same
precedence as subtraction!

So we need another tool for discriminating between operators at the same
precedence level. That tool is 'associativity'. Left-to-right
associativity means "do the stuff on the left first, and use the
calculated result as input for the stuff on the right", and
right-to-left associativity means, of course, the opposite.
Nicely said. However it should be pointed out that in Mathematics
"associative" ordinarily means that it doesn't matter, i.e, if o is an
operator and X, Y, Z are elements, then o is associative if

(X o Y) o Z = X o (Y o Z)

* and + are associative; - and / are not. Left-to-right associativity
and right-to-left associativity are conventions for resolving
unparenthesized expressions.

Mar 16 '07 #3

P: n/a

"Richard Harter" <cr*@tiac.netwrote in message
>
Nicely said. However it should be pointed out that in Mathematics
"associative" ordinarily means that it doesn't matter, i.e, if o is an
operator and X, Y, Z are elements, then o is associative if

(X o Y) o Z = X o (Y o Z)

* and + are associative; - and / are not. Left-to-right associativity
and right-to-left associativity are conventions for resolving
unparenthesized expressions.
And associativity isn't just about notational conventions. It is actually
interesting to mathematicians.

--
Free games and programming goodies.
http://www.personal.leeds.ac.uk/~bgy1mm

Mar 16 '07 #4

P: n/a
On Fri, 16 Mar 2007 21:50:06 -0000, "Malcolm McLean"
<re*******@btinternet.comwrote:
>
"Richard Harter" <cr*@tiac.netwrote in message
>>
Nicely said. However it should be pointed out that in Mathematics
"associative" ordinarily means that it doesn't matter, i.e, if o is an
operator and X, Y, Z are elements, then o is associative if

(X o Y) o Z = X o (Y o Z)

* and + are associative; - and / are not. Left-to-right associativity
and right-to-left associativity are conventions for resolving
unparenthesized expressions.
And associativity isn't just about notational conventions. It is actually
interesting to mathematicians.
Indeed.

Mar 16 '07 #5

P: n/a
On Mar 16, 3:30 pm, c...@tiac.net (Richard Harter) wrote:
Nicely said. However it should be pointed out that in Mathematics
"associative" ordinarily means that it doesn't matter, i.e, if o is an
operator and X, Y, Z are elements, then o is associative if

(X o Y) o Z = X o (Y o Z)

* and + are associative; - and / are not. Left-to-right associativity
and right-to-left associativity are conventions for resolving
unparenthesized expressions.
In mathematics, + and * are associative. Especially in floating-point
arithmetic, they are not (take x = 1e100, y = -1e100, and z = 1, then
(x + y) + z is not the same as x + (y + z). In C, + and * are left
associative, so x + y + z is (x + y) + z.

Mar 17 '07 #6

P: n/a
"christian.bau" wrote:
>
.... snip ...
>
In mathematics, + and * are associative. Especially in floating-
point arithmetic, they are not (take x = 1e100, y = -1e100, and
z = 1, then (x + y) + z is not the same as x + (y + z). In C, +
and * are left associative, so x + y + z is (x + y) + z.
Not "in mathematics". For a single example, consider vector
multiplication. You need to specify the fields involved, rings,
etc.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>

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

Mar 17 '07 #7

P: n/a
In article <45***************@yahoo.comcb********@maineline.net writes:
"christian.bau" wrote:
... snip ...

In mathematics, + and * are associative. Especially in floating-
point arithmetic, they are not (take x = 1e100, y = -1e100, and
z = 1, then (x + y) + z is not the same as x + (y + z). In C, +
and * are left associative, so x + y + z is (x + y) + z.

Not "in mathematics". For a single example, consider vector
multiplication. You need to specify the fields involved, rings,
etc.
In mathematics those are generally not considered standard multiplications.
(Inner product and outer product in 3D, and in higher dimension vectors you
get into tensors when doing outer products.) But '*' is also not associative
in the sedenions. '*' *is* associative in a ring.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Mar 18 '07 #8

P: n/a
On Sun, 18 Mar 2007 00:22:40 GMT, "Dik T. Winter" <Di********@cwi.nl>
wrote:
>In article <45***************@yahoo.comcb********@maineline.net writes:
"christian.bau" wrote:
>
... snip ...
>
In mathematics, + and * are associative. Especially in floating-
point arithmetic, they are not (take x = 1e100, y = -1e100, and
z = 1, then (x + y) + z is not the same as x + (y + z). In C, +
and * are left associative, so x + y + z is (x + y) + z.
Not "in mathematics". For a single example, consider vector
multiplication. You need to specify the fields involved, rings,
etc.

In mathematics those are generally not considered standard multiplications.
(Inner product and outer product in 3D, and in higher dimension vectors you
get into tensors when doing outer products.) But '*' is also not associative
in the sedenions. '*' *is* associative in a ring.
Likewise the octonians.
Mar 18 '07 #9

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