C: Ternary operator giving error when leaving one expression empty

Question

I am attempting to use some code utilising a ternary statement in order to run a specific part, however, when I do this, I get the warning:

expression result unused

And the code in that particular section does not run.

The code in this case is:

i != a ?: printf("|%*s\\\n", i, "");

Why would that be? According to here, this form of the ternary operator, in which there is no alternative for the case, should function, however, it simple skips over it here. Any help is appreciated.

why use a ternary where one part is empty? Wouldn't this be a plain if statement? — Josh Petitt, Feb 23 '13 at 04:11

score 2 · Accepted Answer · answered Feb 23 '13 at 04:12

2

Your code is equivalent to

(i != a) ? (i != a) : printf(...);

Note that you won't end up using the i != a result, hence the warning. Best to write this as an if statement:

if(i==a) printf(...);

answered Feb 23 '13 at 04:12

nneonneo

171,345
36
312
383

Thanks, is there some form however of the ternary statement which is equivalent to: i != a ? printf(...) : printf(...) ? – Bernie Feb 23 '13 at 04:18
Uh, that *is* a ternary statement. – nneonneo Feb 23 '13 at 04:21

C: Ternary operator giving error when leaving one expression empty

1 Answers1