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  
Share this Question
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'. Lefttoright
associativity means "do the stuff on the left first, and use the
calculated result as input for the stuff on the right", and
righttoleft 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.  
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'. Lefttoright associativity means "do the stuff on the left first, and use the calculated result as input for the stuff on the right", and righttoleft 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. Lefttoright associativity
and righttoleft associativity are conventions for resolving
unparenthesized expressions.  
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. Lefttoright associativity
and righttoleft 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  
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. Lefttoright associativity and righttoleft associativity are conventions for resolving unparenthesized expressions.
And associativity isn't just about notational conventions. It is actually interesting to mathematicians.
Indeed.  
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. Lefttoright associativity
and righttoleft associativity are conventions for resolving
unparenthesized expressions.
In mathematics, + and * are associative. Especially in floatingpoint
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.  
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  
P: n/a

In article <45***************@yahoo.com cb********@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/  
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.   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 2566
 replies: 8
 date asked: Mar 16 '07
