operators

**ujwala** · 01-29-2008

Hi i am ujwala i have a confusion regarding inc and dec operators(pre and post).i mean how r they evaluated if they are given in a statemnt.

Code:

#include<stdio.h>
#include<conio.h>
void main()
{
int i=0;
int k;
clrscr();
k=(i++)-(i++)-(--i)-(++i);
printf("%d\n",i);
printf("%d\n",k);
getch();
}

here i calculated in this way plz clear me if i am wrong
now i value is evaluated from left to right
here i got i=2,k=0 which is same as in o/p window

but for this k=(i++)+(i++)+(--i)+(++i);
i got k=2,i=2 but here teh o/p window shows i=2,k=0 how??i dnt understand

**tabstop** · 01-29-2008

You cannot say which way i evaluates. i will get incremented three times and decremented once, for a final value of 2, but the value of k is Indeterminate. The fact that you get a compiled program at all is a miracle in and of itself.

Also: no void main(), but int main().

**Elysia** · 01-29-2008

I hope you actually indent code and this is just a sample with no indentation.
As for question:

Code:

int i = 0;
printf("&#37;d", i++); /* Prints 0, then increments i */
printf("%d", i); /* Prints 1 */
printf("%d", ++i); /* Increments i, then prints 2 */

**Salem** · 01-29-2008

http://c-faq.com/expr/evalorder2.html
The basic problem is, if you modify the same thing twice with side-effects between sequence points (see that FAQ for a definition), then the whole thing is undefined.

At that point, it's useless to try and understand what's going on so you may as well just sit back and admire the light show for what it is.

**matsp** · 01-30-2008

If we break

Code:

k=(i++)-(i++)-(--i)-(++i);

into how it gets evaluated:
k =
(i++)-(i++)
The problem here is that the compiler is allowed to do:
temp1 = i;
i++;
temp2 = i;
i++;
temp3 = temp1 - temp2
or
temp1 = i;
temp2 = i;
i++;
i++;
temp3 = temp1 - temp2

The rest of the calculation has the same problem.

Of course, the result will be different in these two cases. The C standard allows this to be implemented as the compiler likes so that it can be done efficiently by the compiler for "normal cases".

By the way, it would probably be more efficient to write:

Code:

k=(i)-(i+1)-(i)-(i+1); i+=2;

Not to mention that you'd know what is actually going on [and the compiler could reduce it to:

Code:

k = -2; i += 2;

--
Mats

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