c format-specifiers
Updated Sun, 18 Sep 2022 07:26:33 GMT

Format Specifiers in C and their Roles?

wrote this code to "find if the given character is a digit or not"

int main()
    char ch;
    printf("enter a character");
    scanf("%c", &ch);
    printf("%c", ch>='0'&&ch<='9');
    return 0;

this got compiled, but after taking the input it didn't give any output. However, on changing the %c in the second last line to %d format specifier it indeed worked. I'm a bit confused as in why %d worked but %c didn't though the variable is of character datatype.


Characters in C are really just numbers in a token table. The %c is mainly there to do the translation between the alphanumeric token table that humans like to read/write and the raw binary that the C program uses internally.

The expression ch>='0'&&ch<='9' evaluates to 1 or 0 which is a raw binary integer of type int (it would be type bool in C++). If you attempt to print that one with %c, you'll get the symbol table character with index 0 or 1, which isn't even a printable character (0-31 aren't printable). So you print a non-printable character... either you'll see nothing or you'll see some strange symbols.

Instead you need to use %d for printing an integer, then printf will do the correct conversion to the printable symbols '1' and '0'

As a side-note, make it a habit to always end your (sequence of) printf statements with \n since that "flushes the output buffer" = actually prints to the screen, on many systems. See Why does printf not flush after the call unless a newline is in the format string? for details

Comments (1)

  • +0 – Yep, got it! Thanks for clearing the doubt and the tip! — Jul 11, 2022 at 08:11