Output of C++ programs | Set 48 (Bit Manipulation)

Q.1 What Is The Output Of this program ? CPP

#include <iostream>
using namespace std;
int main()
{
    int a = 35;
    int b = 12;
    printf("%d ", ~a);
    printf("%d ", ~ - b);
    return 0;
}

Option a) -36 11 b) -35 -12 c) 220 11 d) 36 11

ans :-a

Explanation:- For any integer n, bitwise complement of n always will be -(n+1). so here -(35+1) and-(-12+1). Q.2 What Is The Output Of this program ? CPP

#include <iostream>
using namespace std;
int main()
{
    int a = 12;
    int b = 25;
    int c = 1;
    printf("%d", a ^ b && c);
    return 0;
}

Option a) 300 b) 1 c) 25 d) 212

ans :-b

Explanation:- In This program, we simple perform a XOR operation between 12(1010) and 25(11001) so result is(10011) equal to the 21, then perform and operation. Q.3 What Is The Output Of this program ? CPP

#include <iostream>
using namespace std;
int main()
{
    unsigned char a = -8;
    unsigned char b = a >> 1;
    printf("%d\n", b);
}

Option a) 128 b) -8 c) -4 d) 124

ans :-d

Explanation:- We know that Negative numbers are represented using 2’s complement of its positive equivalent. So the 2's complement of 8 is (11111000) and we shifted its 1 so its (01111100) is 124. Q.4 What Is The Output Of this program ? CPP

#include <iostream>
using namespace std;
int main()
{
    signed char a = -128;
    signed char b = a >> 1;
    printf("%d\n", b);
    return 0;
}

Option c a) 128 b) 256 c) -64 d) 64

ans :- -64

Explanation:- In this code, we are right shifting -128 by 1. The result will be "-64", binary shift is applied since the operand is a signed value are printed. Q.5 What Is The Output Of this program ? CPP

#include <iostream>
using namespace std;
int main()
{
    unsigned int x = 3;
    x = x | (x << x);
    cout << x;
    return 0;
}

Option a) 27 b) 24 c) 1 d) 0

ans :-a

Explanation:- In this program integer a is 3 its binary (011) shifted 3 its(011000) equal to the 24 and add 3 so its equal to the 27.