?: ternary conditional operator behaviour when leaving one expression empty

Question

I was writing a console application that would try to "guess" a number by trial and error, it worked fine and all but it left me wondering about a certain part that I wrote absentmindedly,

The code is:

#include <stdio.h> #include <stdlib.h>  int main() {     int x,i,a,cc;     for(;;){     scanf("%d",&x);     a=50;     i=100/a; for(cc=0;;cc++) {     if(x<a)     {         printf("%d was too big\n",a);         a=a-((100/(i<<=1))?:1);      }     else if (x>a)     {         printf("%d was too small\n",a);         a=a+((100/(i<<=1))?:1);      }     else     {         printf("%d was the right number\n-----------------%d---------------------\n",a,cc);         break;     } } } return 0; } 

More specifically the part that confused me is

a=a+((100/(i<<=1))?:1);  //Code, code a=a-((100/(i<<=1))?:1); 

I used ((100/(i<<=1))?:1) to make sure that if 100/(i<<=1) returned 0 (or false) the whole expression would evaluate to 1 ((100/(i<<=1))?:***1***), and I left the part of the conditional that would work if it was true empty ((100/(i<<=1))? _this space_ :1), it seems to work correctly but is there any risk in leaving that part of the conditional empty?

Answer 1

This is a GNU C extension (see ?: wikipedia entry), so for portability you should explicitly state the second operand.

In the 'true' case, it is returning the result of the conditional.

The following statements are almost equivalent:

a = x ?: y; a = x ? x : y; 

The only difference is in the first statement, x is always evaluated once, whereas in the second, x will be evaluated twice if it is true. So the only difference is when evaluating x has side effects.

Either way, I'd consider this a subtle use of the syntax... and if you have any empathy for those maintaining your code, you should explicitly state the operand. :)

On the other hand, it's a nice little trick for a common use case.