a[++y]=a[y]+1;

[yiuyuho]

Hello,

can anyone tell me the difference between

a[++y]=a[y]+1;

and

y++;
a[y]=a[y-1]+1;

????
it may be compiler specific, thx in advance

[Dominik Michniewski]

I think that it's compiler specific problem:

Main problem is:
when compiler evaluate right and left side of this expression ...
1. if compiler evaluate right side and after that left -> no difference
2. if compiler evaluate left side and after that right ... no comments

We don't know which strategy will be taken by compiler, so best practice is use second solution

Best regards
DM

[yiuyuho]

hi Dominik! are you just checking post around?

Thanks for your reply, I think I know what you're saying, but isn't '=' always right associative? that's the only way will x=y=z=t=5; work.

[anupam]

I have another problem when i use just like
printf("%d",cas++);
But i get ac. when i use
printf("%d",cas);
cas++;

I don't know why..

[Dominik Michniewski]

anapum:
I always use your first example and I think, that I got Acc correctly

yiuyuho:
hahaha - no, I'm not checking topics around but sometimes I have more time - in other times I check board seldom ...
operator = is right associative , but in your case you have expressions on both sides of =. So compiler may (or not may) evaluate it in any way
i.e
E1 = E2
result of E2 would be used to set E1, but if E1 is expression too - what happened ? E1 will be evaluated first of E2 first ?

Best regards
DM

[Subeen]

I think:
printf("%d", cas++);
and
printf("%d", cas); cas++;
are same.
but
printf("%d", ++cas); is different!

[Ivor]

a[++y]=a[y]+1;

Say y = 1.
++y => y = 2
and your code becomes: a[2] = a[2] + 1

y++;
a[y]=a[y-1]+1;

Same starting point, say y = 1.
y++ => y = 2;
and assignment becomes: a[2] = a[2 - 1] + 1, ie a[2] = a[1] + 1.

That's the difference. I haven't tested it but it is the way I have always understood C compiler. Shame, if I'm wrong. Usually if I'm not sure about the statement, I simply rewrite it. Better be clear.

[CDiMa]

Hello,

can anyone tell me the difference between

a[++y]=a[y]+1;

and

y++;
a[y]=a[y-1]+1;

????
it may be compiler specific, thx in advance

By the C standard the first assignment is undefined. That means that anything may happen, even that your program solves all problems in all ACM volumes at once

Seriously, the ++ operator has a double meaning: it evaluates as an expression and has the side effect of changing the value of y. Side effects are guarateed to be complete at the next sequence point. Since in

Code: Select all

a[++y]=a[y-1]+1;

the sequence point is at the semicolon, you can't establish if the side effect has taken effect while accessing y to compute y-1. Furthermore if a variable is written to in an expression, all accesses to it within the same expression are allowed only to compute the value to be written. In your case you are using the the value of y to evaluate y-1 and not the value to be written to y.

Hope this clears things a little bit...

Ciao!!!

Claudio

[anupam]

I think:
printf("%d", cas++);
and
printf("%d", cas); cas++;
are same.
but
printf("%d", ++cas); is different!

But subeen, I faced it in a problem names
monkey in a forest or like that..

you corrected my fault as i can remember. Again, I faced it during the conest of dhaka regional (Cyber Cafe, the problem)..
--
May be I am wrong.
or would you just check your previous posts??

[CDiMa]

a[++y]=a[y]+1;

Say y = 1.
++y => y = 2
and your code becomes: a[2] = a[2] + 1

Not quite right...
++y results 2, value of y cannot be read for anything else until ';'
your code becomes: ?

That's the difference. I haven't tested it but it is the way I have always understood C compiler. Shame, if I'm wrong. Usually if I'm not sure about the statement, I simply rewrite it. Better be clear.

Sequence points are an obscure matter... better be clear

Ciao!!!

Claudio

[little joey]

i=1;printf("(%d,%d)",i++,i++);
can write (1,1) (1,2) (2,1) or even (2,2)
but i=1;(printf("%d",(i++,i++));
should write 2

[CDiMa]

i=1;printf("(%d,%d)",i++,i++);
can write (1,1) (1,2) (2,1) or even (2,2)
but i=1;(printf("%d",(i++,i++));
should write 2

i++,i++ isn't tricky and is well defined since the comma is a sequence point

Ciao!!!

Claudio

[yiuyuho]

Thank you very much for your explainations everyone, I guess it is complier specific.

I'll be more careful in the future about these kinds of stuff, thx a lot!

[horape]

Hello,

can anyone tell me the difference between

a[++y]=a[y]+1;

and

y++;
a[y]=a[y-1]+1;

????
it may be compiler specific, thx in advance

First one is invalid (unspecified/undefined really, don't know for sure), the order in which the ++y on the left side and the y on the right one are evaluated is not defined. You can not trust what will that code do.

More on http://www.eskimo.com/~scs/C-faq/q3.2.html

Saludos,
HoraPe

[Farqaleet]

Following is what MSDN says

"In the prefix form, the increment or decrement takes place before the value is used in expression evaluation..."

By this the increment is done before y is used anywhere in the expression. That's what microsoft says. but then again you can't say anthing otherwise without checking other docs eg. Borland, gcc, g++ etc.