How to find the value of 2^n, without using semicolon anywhere in program & value of n is defined by user?

Question

please suggest the most time efficient way...

Answer 1

Interesting Baptiste! Anyway I find it's a strange requirement.. 😅

Answer 2

#include <stdio.h>
#include <stdlib.h>

int main( int argc, char** argv )
{
if( argc > 1 && printf( "%d", 1 << atoi( argv[1] ) ) ){}
}

No semicolons, but you have to run your program with an argument like:

program 12

output: 4096

Answer 3

I could not avoid 2 semicolon whether using a function or not due to the user input problem and return statement

#include <stdio.h>

int call(unsigned n,unsigned res){
if(scanf("%d",&n) && (res=1)){
while(n-- && (res<<=1)){}
}

return res;
}

int main() {
if(printf("%d\n",call(0,0))){}
return 0;
}