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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();
int j = g() | (f() + g());
}
int f()
{
if (x == 0)
return x + 1;
else
return x – 1;
}
int g()
{
return x++;
}

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();
int j = g() | (f() + g());
}
int f()
{
if (x == 0)
return x + 1;
else
return x – 1;
}
int g()
{
return x++;
}
a) i value is 1 and j value is 1
b) i value is 0 and j value is 0
c) i value is 1 and j value is undefined
d) i and j value are undefined

Answer: c

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