Question

#include<stdio.h>

main()
{
       int x = 5, y = 10, z = 10;    

       x = y == z;          // This computational expression causes the value of x to be 1. I fail to understand why
       printf("%d\n", x); //Why is the value of x 1 here.

}

I fail to understand the statement x = y ==z;

According to me – x = 10 since y == z. z=10 and is stated to be equivalent to y. The value of y is then assigned to x – x = y

Answer 1

You assign the result of the comparison »Is y equal to z« to x, which is 1, i.e. true.

Note the different operators:

x = ... // assignment
y == z  // comparison with either 0 (false) or 1 (true) result

Let’s break the program down a little bit further:

1. You initialize x to 5 and y and z to 10.
2. You perform a comparison (see above) of y and z without caring for the result. So that’s a line that can safely be ignored. But it results in 1 since y is equal to z.
3. You print the current values of all three variables.
4. You perform the same comparison, this time assigning the result to x. x now has the value »Is y equal to z«, which is 1 in this case.