about ++ operator

[virus]

I have Red Hat Linux enterprise-4 edition on my pc. Problem is this.


int main()
{
int i=1;
i=i++;
printf("%d",i);
return 0;
}
Out put of this program is 1 in my pc & 2 in turbo c. what is the correct result.
Please help me.

[horace1]

I have Red Hat Linux enterprise-4 edition on my pc. Problem is this.


int main()
{
int i=1;
i=i++;
printf("%d",i);
return 0;
}
Out put of this program is 1 in my pc & 2 in turbo c. what is the correct result.
Please help me.

The value of i after
i=i++;
depends on whether the increment of i occurs before or after the assignment. in C it's unspecified; in some implementations the value will be 1, in some it will be 2. Simillarly
(i * j) + (i * j++)
will the value of j in (i * j) be that before or after the increment in (i * j++) ? again it is unspecified.
The advice is do not use a variable more than once in an expression if one or more of the references has a ++ or -- operator attached to it.

[Ganon11]

Indeed. The statement

i=i++;

is like saying

i = i = i + 1;

Why not just say

i++;

and be done with it? There's no need to reassign i to itself, since ++ stands for

i = i + 1;

[horace1]

have a look at the C FAQ on URL
http://c-faq.com/

in particular the discussion about "i=i++;" on URL
http://c-faq.com/expr/ieqiplusplus.html

and " will a[i] = i++; work" on URL
http://c-faq.com/expr/evalorder1.html

there is also a good discussion on expressions and "side effects" on URL
http://www.csci.csusb.edu/dick/c++std/cd2/expr.html

where it states in paragraph 4
"4 Except where noted, the order of evaluation of operands of individual
operators and subexpressions of individual expressions, and the order
in which side effects take place, is unspecified. ....."