include int main int i 4 int j 8 return 0 include int main

#include int main () int i 4; int j 8 return 0; # include int main() int i= 32, j =0x20, k, 1, m: m k^1; = printf(\"%d, %d, td, td\ \", j, k, l, m); return 0

Solution

Here is the explanation for the first 2 problems:

#include <stdio.h>
int main()
{
int i = 4;   //4 in binary = 0000 0100
int j = 8;   //8 in binary = 0000 1000
printf(\"%d, %d, %d\ \", i|j&j|i, i|j&&j|i, i^j);
return 0;
}

Here the operators used are: |, &, &&, ^.
And the precedence is: &, ^, |, &&
Therefore,
i|j&j|i = i|(j&j)|i
= i|j|i
= (i|j)|i
= (0100 | 1000) | (0100)
= 1100 | 0100
= 1100 = 4 + 8 = 12.

i|j&&j|i = (i|j)&&(j|i)
       = (0100 | 1000) && (1000 | 0100)
       = 1100 && 1100
       = 12 && 12 = 1. Here 1 holds true, and 0 holds false.

i^j = 0100 ^ 1000 = 1100 = 12.
Therefore, the output is: 12, 1, 12.

#include <stdio.h>
int main()
{
int i = 32, j = 0x20, k, l, m;   //32 in binary = 0010 0000, 0x20 in binary = 0010 0000.
  
k = i|j;  
l = i&j;
m = k^l;
  
printf(\"%d, %d, %d, %d\ \", j, k, l, m);
return 0;
}

i|j = 0010 0000 | 0010 0000 = 0010 0000 = 32(k).
i&j = 0010 0000 | 0010 0000 = 0010 0000 = 32(l).
k^l = 0010 0000 ^ 0010 0000 = 0000 0000 = 0(m).

Therefore, the output is: 32, 32, 32, 0