C中三元运算符中的条件表达式

Question

Code

#include <stdio.h>

int main() {
  int i;
  for (i=1; i<=10; i++) {
        (i % 2) ? printf("%d is odd\n", i) : printf("%d is even\n", i);
  }
}

Result

1 is odd
2 is even
3 is odd
4 is even
5 is odd
6 is even
7 is odd
8 is even
9 is odd
10 is even

In the above C program, why it still works fine even though the conditional expression only states i%2 and not i%2!=0 ?

Answer 1

In C, integers can be used in a Boolean context, and zero represents false while non-zero represents true.

That's why your code works. The expression num % 2 will be 0 (the single false value) for an even number and 1 (one of the many possible true values) for an odd number.

The following expressions would all work for detecting an odd number:

num % 2
(num % 2) != 0
((num % 2) != 0) != 0
... and so on, ad untilyougetboredum (like 'ad infinitum' but with limits).

Having said that, I don't really consider it a good idea to do it this way, code should express intent as much as possible and the intent here should be to choose the path of execution based on a comparison. That means, if you're looking for an odd number, you should use something like (num % 2) == 1 .

You also don't need a separate printf call in each of those code paths:

printf("%d is %s\n", num, ((num % 2) == 1) ? "odd" : "even");

---

You'll notice I've also used num instead of i . This is simply a style thing of mine, related to the afore-mentioned intent. If the variable is only used as an index, I'm happy to use the i -type variables ^(a) but, the second it gains a semantic property (like a number being checked for oddity), I tend to use more descriptive names.

I have no issue with people using simple variable names, I just prefer more descriptive ones in my own code.

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^(a) Actually, I'd probably use idx in that case but that's being too CDO ^(b) , even for me :-)

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^(b) OCD but in the right order :-)

Answer 2

C doesn't have a dedicated boolean type. It uses int value as boolean. That is 0 is considered false and any non zero value is treated as true . Try printing some conditions

printf("%d",5==5);
printf("%d",1>3);

This will output 1 and 0.

C always uses 1 to denote true . But any other non-zero value would work as well when using in conditions.

if(6+1)
    printf("TRUE");

Will print TRUE.

This is also the reason we can use this form of while loop:

int i= 10;
while(i--){
    printf("%d",i);
}

Will print 9876543210 . Notice it stops when i becomes 0, which is false .

Now back to the question, i%2 would always result in either 0 or 1. In case of 1(true) the first statement is run while in case of 0 (false) the second statement is run.