Explain the given C program.

Question

<code>#include <stdio.h>
        int main()
        {
            unsigned int b = 20;
            b = ~b;
            printf("%d\n", b);
        }</code>

Comments

updated.

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yes, it will be -21...

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I also know the answe. can you ecplain ?

Answer 1

unsigned int b=0000000000010100 (assuming int = 2 bytes)

b=~b, therefore b=1111111111101011

now %d specifier does not know anything about signed or unsigned, it just assume that the given number is in signed 2's complement form..

and hence value returned will be -21...

Comments

One question. Why will it return in 2's complement format?

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@srestha %d specifier assume that given input is in 2's complement....no matter you provide it in either unsigned char or signed char or unsigned int or signed int...it will always expand it assuming 2's complement

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ok say, code is like this

#include <stdio.h>
        int main()
        {
            int b = 20;
            printf("%d\n", b);
        }

will the b value be print in 2's complement format?

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20...whats problem in this??

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u have told in ur answer " %d specifier does not know anything about signed or unsigned, it just assume that the given number is in signed 2's complement form.."
So why not in above program it is in 2's complement form?

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see this, (assuming int of 2 bytes)
int b=20;
in 2's complement its value will be b= 0000000000010100..
now %d will evaluate it(0000000000010100) assuming 2's complement and the value printed will be 20 because 0000000000010100 is a value of 20 itself in 2's complement..

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I still cant understand , how it comes to be -21, ??????