called object is not a function or function pointer in use of ternary

Question

I am constantly receiving:

error: called object is not a function or function pointer

When using ternary operator like that:

puts("\nx: " (0 == 1) ? "y1\n" : "y2\n");

---

What am I doing wrong?

Answer 1

You are attempting to call an object which is neither a function nor a function pointer! In particular, the compiler is seeing the open paren after the string and thinks (as much as a compiler can be said to "think") that you are trying to invoke a function call. You cannot concatenate a string with the ternary operator as you are trying to do. Try:

printf("\nx: %s", (0 == 1) ? "y1\n" : "y2\n");

Answer 2

puts("\nx: " (0 == 1) ? "y1\n" : "y2\n");

This is not the right way to do what you want, because you cannot concatenate C strings like this.

You can do this using puts():

puts("\nx: ");
puts((0==1) ? "y1\n" : "y2\n");

EDIT: (Suggested by @WilliamPursell

To avoid the appending of an unwanted newline character, use fputs() instead of puts().

fputs("\nx: ");
fputs((0==1) ? "y1\n" : "y2\n");

Answer 3

You can not concat the strings the way you do.

The simple solution is to use printf

printf("\nx: %s", (0 == 1) ? "y1\n" : "y2\n");

or if you insist on using puts you need the strcat function from string.h

char s[256] = "\nx: ";
puts(strcat(s, (0 == 1) ? "y1\n" : "y2\n"));

Answer 4

This is your code

"\nx: " (0 == 1) ? "y1\n" : "y2\n"
/*     ^ this is ignored        */

so it seems as if the string literal was being called as a function -> "\nx: "(0 == 1), doesn't it look like that?

You can achieve what you want with the printf() function like this

printf("\nx: %s\n", (0 == 1) ? "y1" : "y2");