How this code execute ?

Question

#include <stdio.h>
int main()
{
int n=2,m=2;
while(m--)
n*=n;
printf ("%d",n);
return 0;
}

Answer 1

Let me translate everything inside main() to english.

1) create 2 variables, one with name 'n'(initial value = 2) and another with name 'm'(initial value = 2).

2) loop till the time 'm' has a non-zero value (zero in C defaults to false) and decrement it after every iteration

3) inside the loop, square the value present in variable 'n' . So values inside loop would be
For m == 2
n = 2*2 == 4
For m == 1
n = 4*4 == 16
Now value of 'm' is zero so break the loop.

4) print the value inside 'n' on the console which is 16

Answer 2

Now i get it
Thanks buddy.
Arsenic

Answer 3

Since, the value of m is 2. So, loop is executed 2 times because value of m is decreases by 1 after completing 1 loop.

In first round n = 2*2 = 4.
In second round n = 4*4 = 16.

So, in output screen compiler shows 16.

https://code.sololearn.com/cvZ6x2Kc7s7r/?ref=app

Answer 4

Thanks Ashish Gupta