What does "%. * S" mean in printf?

Question

I have a piece of code in which there is

printf("%.*s\n") 

what does %.*s mean?

Answer 1

You can use an asterisk ( * ) to convey the specifier / precision of the width of printf() , rather than hard-coded it into a format string, i.e.

 void f(const char *str, int str_len) { printf("%.*s\n", str_len, str); } 

Answer 2

I do not think the code above is correct, but (according to this description printf() ) .* Means

The width is not indicated in the format string, but as an additional argument of an integer value preceding the argument to be formatted. ''

So this is a string with a valid width as an argument.

Answer 3

See: http://www.cplusplus.com/reference/clibrary/cstdio/printf/

.* Accuracy is not indicated in the format string, but as an additional argument of an integer value preceding the argument to be formatted.

s character string