What is happening in the if condition?

Question

One of my friends ask me about this code and I and he are unable to find out, what is happening in the if condition. Can you guys explain how this condition works?

int main()
{
   int i;

   if (i = (1, 2, 0))
       printf("Mehrose");
   else
       printf("Hello ");

   printf("%d\n", i);

   return 0;
}

The output for this code is Hello 0

You might want to learn about [the comma operator](https://en.cppreference.com/w/c/language/operator_other#Comma_operator) and how it works. — Some programmer dude, Nov 06 '19 at 15:11

Reticulated Spline · Accepted Answer · 2019-11-06T16:00:45.497

First, formatting as the compiler sees the code we get:

int main(void)
{
   int i;
   if(i=(1,2,0))
      printf("Mehrose");
   else 
      printf("Hello");
   printf("%d\n",i);
   return 0;
}

The if statement can be broken down:

Comma operator , is evaluated first, left-hand side of the operator is discarded. This repeats for each operator:

if(i=(1,2,0))

if(i=(2,0))

if(i=0)
The assignment operator = assigns a value of 0 to i, and returns the right-hand side of the expression:

if(0)
Recall that 0 is evaluated as false (is "falsy") and 1 is evaluated as true (is "truthy"). Thus the first condition fails and the second block is executed. "Hello" is printed to the standard output stream, followed by "0".

score 2 · Answer 2 · answered Nov 06 '19 at 15:19

In the expression,

i=(1,2,0)

you're using the comma operator, which evaluates all its operands and yields the result of its rightmost operand - which is 0 here. So 0 is assigned to i.

So it's equivalent to if (i = 0), which assigns 0 to i and yields the value of i which is false and thus it prints the string in the else branch.

What is happening in the if condition?

2 Answers2