C PROGRAMMING

Question

OUTPUT OF THE FOLLOWING PROGRAMME

#include<stdio.h>
int main()
{
    int n1=10;
    int n2=15;
    int n3;
    n3=~n1;
    printf("%d",n3);
    return 0;
}

Basically here i want to understand how do we store negation value and what are the changes done,for a signed and unsigned integers here considering integer is of 4 bytes here. 

Answer 1

The complementation operation being used here is like NOT gate in digital logic . So the functionality is same as that of NOT gate.

Hence when we do the complementation of a number , the bits of the original number are toggled . 

Hence binary representation of 10 in 32 bits : 28 0's followed by "1010"..

So by doing complementation we get  28 1's followed by "0101"..

Now we know :

Numbers are stored in 2's complement representation in computer for sake of simplicity of its usage in arithmetic circuits and unique representation of '0' as well..So 

a) If MSB is 0 , the number itself gives true magnitude.

b) If MSB is 1 , we need to take 2's complement of that number to get true magnitude and then add '-' sign to it.

So in this case after complementation we have 28 1's followed by "0101" . So we need to take 2's complement to get true magnitude of the number..

Taking 2's complement we have : 28 0's followed by "1011"  ==>  -11

Hence we obtain -11 as the output of the program..

P.S. : There is difference between '!' and '∼' operator . The first one is "logical not operator also known as negation" whereas second one is bitwise NOT known as complementation operator. 

Comments

yes. But 2's complement is not mandated by C, and this is implementation detail and thus this question becomes out of GATE scope.

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Yes thats true bt I think most of the systems use 2's complement representation by default..