Please give me the output and explain it

Question

#include<stdio.h>
int main()
{
static int i;
for(++i;++i;++i)
{
printf("%d",i);
if(i==4)
break;
}
return 0;

}

Answer 1

You do realise it's a 2 and a 4 not 24.

Answer 2

Output is 24
but why ?

Answer 3

I think so but how it is possible

Answer 4

The i starts off with a value 0;
In the for loop:
In the first section of the for loop.....i is incremented...so i = 1.
Then the second portion is evaluated for a true of false...but i is incremented again in there...now i = 2...so the condition is true..i.e not 0.
The action statements in your for loop are then executed.....output = 2
then the i is incremented in the third portion...i = 3.
Then...the condition (the second portion) is evaluated again....but i is incremented again in there....i = 4....now it prints 4 and breaks because of the if test is true.

Answer 5

I think you forgot one step after i=3 it's increment in first step hence i=4 and in 2nd it check and increment i=5

Answer 6

Please explain again