C - Output explanation of printf("%d %d\n",k=1,k=3); [duplicate]

Question

How to explain the output of the below code:

include <stdio.h>

int main(void) {
    int k;
    printf("%d %d\n",k=1,k=3);
    return 0;
}

Ideone Link

My thinking was that 1 will be assigned to k variable and then 1 would be printed. Similarly 3 will be assigned to k and output will be 3.

Expected Output

1 3

Actual Output

1 1

I am extrapolating from

int a;
if (a = 3) { 
    ...
} 

is equal to

if (3) { 
    ... 
}

Please let me know where am I going wrong?

Answer 1

The problem is, the order of evaluation of function arguments are not defined, and there's no sequence point between the evaluation or arguments. So, this statement

 printf("%d %d\n",k=1,k=3)

invokes undefined behavior, as you're trying to modify the same variable more than once without a sequence point in between.

Once a program invoking UB is run and (if) there's an output, it cannot be justified anyway, the output can be anything.