why the output in negative (using bitwise-not)

Question

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?

int's size must >= 16 bit, char >= 8 bit. There's no 4-bit data type in C — phuclv
– phuclv, Commented Mar 7, 2014 at 9:31

brokenfoot · Accepted Answer · 2014-03-07 09:25:00Z

2

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

answered Mar 7, 2014 at 9:25

brokenfoot

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Comments

Guffa · Accepted Answer · 2014-03-07 09:22:06Z

1

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.

answered Mar 7, 2014 at 9:22

Guffa

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Comments

timrau · Accepted Answer · 2014-03-07 09:22:37Z

1

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

answered Mar 7, 2014 at 9:22

timrau

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jvm1990 · Accepted Answer · 2014-03-07 09:23:03Z

0

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

answered Mar 7, 2014 at 9:23

jvm1990

1

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