5 replies to this topic

[qwertyuiop]

what is the effect of fllowing code n values of a and b
b = 0;
a = 1;
b = a++ + a++;

[v0id]

The result will be:

a = 3
b = 2

When you're using the post-increment-operator, the value of the variable will be incremented, but it will return the old value. In the following code, the variable will have the incremented value.

c = 10
d = c++
// c = 11
// d = 10

[WingedPanther]

Void, have you tested that? (I haven't) My logic was the following:

Expanding on that:
b = a++ + a++
the first a++ increments a to 2, returns 1, leaving an intermediate calculation of:
b = 1 + a++
the remaining a++ increments a to 3, returns 2, leaving an intermediate calculation of:
b = 1 + 2

Result: a = 3, b=3.

[v0id]

Actually, I made something like you did, but before I posted it, I tested it using GCC, and I got kinda surprised. The result was as I wrote in my post. For some reason does neither of the a's increment more than one, before the statement has ended. It could be bug - I don't know.

This is the little C-program I used for testing.

#include <stdio.h>

int main()
{
        int a = 1;
        int b = 0;

        b = a++ + a++;
        printf("a: %d\nb: %d\n", a, b);

        return 0;
}

gcc test.c -o test

$ ./test
a: 3
b: 2

[G_Morgan]

i++ is post increment so should increment the value after the statement has been executed.

Try ++a and it should then give a = 3, b = 6;

I wouldn't actually write any code like that. It's the reason a lot of people dislike C.

//edit - just thought I'd add what the compiler will convert these things to

[HIGHLIGHT="C"]
b = a++ + a++;
/*converts to */
b = a + a;
a += 1;
a += 1;

-------------
b = ++a + ++a;
/*converts to*/
a += 1;
a += 1;
b = a + a;
[/HIGHLIGHT]