If unsigned types should not have negative values, why is it negative when I turn on all the bits?

Question

#include <stdio.h>
#include <math.h>
#include <limits.h>

int main(void)
{
    unsigned long x = 0;

    x = x ^ ~x;
    printf("%d\n", x);

    x = (unsigned long)pow(2, sizeof(x)*8);
    printf("%d\n", x);

    x = ULONG_MAX;
    printf("%d\n", x);
    return 0;
}

I am using CodeBlocks12.11, and MinGW 4.7.0-1 on Windows 7. And for some reason I am having trouble making my variable x acquire the largest possible decimal value representation. Why does this happen, I am sure that x = ULONG_MAX should work but it also results in -1, now surely that is not right! I tried compiling it outside of Code-Blocks as well.

What am I missing here?

Yes, as @Zoidberg says, the compiler's warning would have pointed out your mistake right away. — vonbrand
– vonbrand, Commented Mar 30, 2013 at 3:15

score 9 · Accepted Answer · 2013-03-30 00:51:59Z

9

You have to print unsigned variables with u. A long is prefixed with l, hence you need lu in this case.

printf("%lu\n", x);

edited Mar 30, 2013 at 0:51

answered Mar 30, 2013 at 0:40

user1944441

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8 Comments

Leonardo Over a year ago

Thank you, I knew it would be something terribly obvious!

user1944441 Over a year ago

@PascalCuoq Tnx made a typo.

Mysticial Over a year ago

Isn't that supposed to be %lu?

Mysticial Over a year ago

@Armin I didn't downvote, but that's not correct. It's possible for unsigned long to be larger than unsigned. It might not be on your system, but the standard allows it.

user142019 Over a year ago

@Leonardo turn on compiler warnings and listen to them. Any modern compiler will complain when the format specifiers do not match the types. You must not use the wrong specifier since that will (and should) undoubtedly break sometime.

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