Howz this equality operator working?

Code:

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#include<stdio.h>
main()
{
        int k=35;
        printf ("%d %d %d", k==35, k=50, k>25);
       
}

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Code:

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Output:
0 50 1

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I thought it was just comparing. Like it would return a 1 if k was 35 and a 0 if k was not not 35! But as it happens, even though I change the value of k, I can't see any change in the output. (It still remains 0)

Please clarify. :)

I'm not a pro C programmer, I'm more of a C++ programmer, but if I remember, printf() uses va_lists to let you have as many parameters as you want and these va_lists are a bunch of macros, if I recall. You might want to surround your expressions with parentheses.

Maybe fix your program?
http://c-faq.com/expr/seqpoints.html

You have NO control over when the assignment happens, and when the comparison is performed between the two sequence points which bound that particular statement.

Quote:

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Originally Posted by Salem

Maybe fix your program?
http://c-faq.com/expr/seqpoints.html

You have NO control over when the assignment happens, and when the comparison is performed between the two sequence points which bound that particular statement.

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Still not sure what that means!

Well, this was an assignment problem and I blew my brains out pondering over this thing.

Please explain how I don't have control when the assignment happens. :)
Just wanted to know how printf sees the equality operator.

In the same way that
printf( "%d %d\n", i++, i++ );
is undefined.

Yours may degenerate into

k = 50;
printf ("%d %d %d", k==35, k, k>25);

or

printf ("%d %d %d", k==35, k, k>25);
k = 50;

Both are correct as far as valid C code is concerned, and the compiler writer is free to choose which one is best given the circumstances. The problem for you is that when you start pushing on the assumptions of the behaviour, sometimes you get bitten.

Remember, C is a compiled language, so all sorts of things happen between sequence points. What most certainly doesn't happen is the code is executed in the strict L->R order like it might be by a simplistic interpreter.

Fixing a couple of trivial problems, we get

Code:

---

#include<stdio.h>
int main()
{
        int k=35;
        printf ("%d %d %d", k==35, k=50, k>25);
        return 0;
}

$ gcc -W -Wall -ansi -pedantic -O2 foo.c
foo.c: In function `main':
foo.c:5: warning: operation on `k' may be undefined
foo.c:5: warning: operation on `k' may be undefined

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Quote:

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Originally Posted by Salem

printf ("%d %d %d", k==35, k, k>25);
k = 50;

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You mean

Code:

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printf ("%d %d %d", k==35, 50, k > 25);
k = 50;

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Quote:

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Originally Posted by Rashakil Fol

You mean

Code:

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printf ("%d %d %d", k==35, 50, k > 25);
k = 50;

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that's also legal.