CODEWITHSUNDEEP
cformat-specifiers
Just Zac
Just Zac

Reputation: 43

format specifier value issue in c

int friends = 20;

printf("I have %d friend%c", friends , (friends !=1 ? "s" : ""));

return 0;

So whenever I run the code it debugs into this

I have 20 friend$

It works fine when I run it with %s format specifier after the friend. s is only one character so why doesn't it work?

Upvotes: 0

Views: 311

Answers (2)

Tauqeer
Tauqeer

Reputation: 81

"s" is not a one character, its a String Literal whose type is char *
's' is one character Constant whose type is int.
String Literal in C requires double quotes, and printf( ) have format specifier %s to print string literals.
Character Constant in C requires single quotes, and printf( ) have format specifier %c to print them.

Now your code snippet, 

printf("I have %d friend%c", friends , (friends != 1 ? "s" : ""));

will return string literal, since

"s" or " "

is string literal and printf( ) requires format specifies "%s" to print them.

if you want to use %c format specifier in code snippet, then use, character constant 's' instead of string literal "s"

printf("I have %d friend%c", friends , (friends != 1 ? 's' : ' '));
                                                              ^

Also note that, there must be a space between single quotes as show above the caret symbol otherwise it would give error: empty character constant.
In case of string literals, empty string is allowed.

Upvotes: 1

Achal
Achal

Reputation: 11921

so why doesn't it work? because %c expects char but expression (friends !=1 ? "s" : "") results in strings(double quotes). So either use %s like

printf("I have %d friend%s", friends , (friends !=1 ? "s" : ""));

Or

Replace "s" with 's' and "" with ' ' as %c expects char.

printf("I have %d friend%c", friends , (friends !=1 ? 's' : ' '));

Upvotes: 3

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