Question

#include<stdio.h>
int main()
{
int k=35;
printf("\n %d %d %d",k==35,k=50,k>40);
return 0;
}

In this program k is already assigned the value 35.what will be the output sir. Please explain it.

Answer 1

The output of the program is not defined by standard. Because output of this program depends on the order in which the function arguments are evaluated. In this case “,” is a separator of function arguments and this is not a sequence point.