C equivalent operator creating confusion

Question

Program 1:

#include<stdio.h>
void main()
{
    int i=55;
    printf("%d %d %d %d\n",i==55,i=40,i==40,i>=10);
}

Program 2:

#include<stdio.h>
void main(){
    int i = 55;
    if(i == 55) {
        printf("hi");
    }
}

First program give output 0 40 0 1 here in the printf i == 55 and output is 0 and in the second program too i ==55 and output is hi. why

Answer 1

In the first example, the operators are evaluated in reverse order because they are pushed like this on the stack. The evaluation order is implementation specific and not defined by the C standard. So the squence is like this:

1. i=55 initial assignment
2. i>=10 == true 1
3. i==40 == false (It's 55) 0
4. i=40 assignment 40
5. i==55 == false (It's 40) 0

The second example should be obvious.