
#include<stdio.h>
main()
{
char*p;
printf("%d%d", sizeof (*p), size of(p)) ;
}
int f(int x, int *py, int **ppz)
{
int y, z;
**ppz += 1;
z = **ppz;
*py += 2;
y = *py;
x += 3;
return x + y + z;
}
void main()
{
int c, *b, **a;
c = 4;
b = &c;
a = &b;
printf("%d ", f(c, b, a));
return 0;
}

What will be the final values of i and j in the following C code?#include <stdio.h>int x = 0;int main(){int i = (f() + g()) | g(); //bitwise orint j = g() | (f() + g()); //bitwise or}int f(){if (x == 0)return x + 1;elsereturn x - 1;}int g(){return x++;}a) i value is 1 and j value is 1b) i value is 0 and j value is 0c) i value is 1 and j value is undefinedd) i and j value are undefined
#include <stdio.h>What is the output of this program?void solve) {int x = printf ("Hello") ;printf(" %d", x);}int main(){solve() ;return 0;}