CODEWITHSUNDEEP
cbit-manipulationbitwise-operators
Itay Avraham
Itay Avraham

Reputation: 393

why the output in negative (using bitwise-not)

void main()
{
        unsigned int a = 10;
        a = ~a;
        printf("%d\n", a);
}

the output is -11

10 = 1010

~10 = 0101

why the output is negative?

Upvotes: 0

Views: 137

Answers (4)

brokenfoot
brokenfoot

Reputation: 11609

Use %x to view the consistent hex result.

#include <stdio.h>
void main()
{
        unsigned int a = 10;
        printf("%x\n", a);

        a = ~a;
        printf("%x\n", a);
        return 0;
}

Output:

a
fffffff5

Upvotes: 2

jvm1990
jvm1990

Reputation: 1

use %u instead of %d because printf treats your variable according to %d or %u. u for

Upvotes: 0

timrau
timrau

Reputation: 23058

%d is for signed decimal integer. Use %u to print an unsigned integer in decimal.

Upvotes: 1

Guffa
Guffa

Reputation: 700152

The result of ~1010 is not 0101 but 11111111111111111111111111110101. All 32 bits of the value are reversed, not only the bits up to the highest set bit.

As the 32nd bit is set in the result, it's negative.

Upvotes: 1

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