conditional operator usage using printf in C

Question

I need to consider the user input character is white space or not. The program works well when I type in a blank space, * is printed out successfully. But when I type a character that is not white space, I can't get %. Instead the character that I entered is just printed out.

Is there a problem with my conditional operator code?

This is my code:

#include <stdio.h>
#include <ctype.h>

int main() {
    char character;
    printf("Press any single key\n");
    character = getchar();
    
    (isspace(character) > 0) ? printf("*") : printf("%");

    return 0;
}

`printf("%")` should be `printf("%%")`. A percent character on its own needs to be "escaped", because a complete format instruction such as `"%d"` is expected. — Weather Vane, Oct 01 '22 at 13:23
Fwiw printf(isspace(character)?”*”:”%%”); would work also. — Jeremy Friesner, Oct 01 '22 at 13:25
Does this answer your question? [How to escape the % (percent) sign in C's printf](https://stackoverflow.com/questions/1860159/how-to-escape-the-percent-sign-in-cs-printf) — phuclv, Oct 01 '22 at 22:40

chqrlie · Answer 1 · 2022-10-01T13:34:19.450

% is a special character in a printf format string. To output a % characters, you can:

either use printf("%%");
or use putchar('%');

Also note these problems:

character should be defined with type int to reliably store all return values of the function getchar(), including the special negative value EOF.
isspace() is defined for values of type unsigned char and the special value EOF returned by getchar(), do not pass a char value. Instead, define character as an int and pass that.
isspace() does not necessarily return a positive value for whitespace characters, you should just test if the return value is non zero, which in C can be written:
```
  if (isspace(character)) {
      ...
```

Here is a modified version:

#include <ctype.h>
#include <stdio.h>

int main() {
    int c;

    printf("Press any single key\n");
    c = getchar();
    
    if (isspace(c)) {
        putchar('*');
    } else {
        putchar('%');
    }
    return 0;
}

conditional operator usage using printf in C

1 Answers1