Why is the output "false2", when the condition at the if block is 1 which is true?

Question

Why is the output "false2", when the condition at the if block is 1 which is true ?

#include<stdio.h>
int main(){
int x=0;
if (x++)
printf("true1");
else if (x==1)
printf("false2");
return 0;}

Answer 1

when the condition at the if block is 1 which is true ?

That's not the case. The if condition is actually false.

The expression x++ yields the current value of x (which is what's used to test the condition) and then increments it. Since x's current value is 0, it evaluates to false and thus it goes into the else if condition.

Contrast this with ++x which increments first and then yields the new value.

See C: What is the difference between ++i and i++?

Answer 2

x++ means first use the value of x and then increment it's value by 1. So your if condition is if(0), Then x value increments and becomes 1. So your else if condition is else if(1). Whereas ++x means first increase the value of x by 1 and then use it so your first condition is if(1).

Answer 3

The first time x is tested, it is zero and then it is incremented:

//x is post-incremented, i.e. evaluated before being incremented
int x=0;
if (x++)//x is tested when x==0, then is incremented
printf("true1");
else if (x==1)//execution flow goes here...
printf("false2");//...then here

To make x test as TRUE on the first iteration change it to a pre-increment:

//x is pre-incremented, i.e. incremented before being evaluated
int x=0;
if (++x)//x is incremented, then tested when x==1
printf("true1");//execution flow goes here...
else if (x==1)
printf("false2");
//...then here