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

Question

I got a code snippet in which there is a

printf("%.*s\n") 

what does the %.*s mean?

Answer 1

You can use an asterisk (*) to pass the width specifier/precision to printf(), rather than hard coding it into the format string, i.e.

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

Answer 2

More detailed here.

integer value or * that specifies minimum field width. The result is padded with space characters (by default), if required, on the left when right-justified, or on the right if left-justified. In the case when * is used, the width is specified by an additional argument of type int. If the value of the argument is negative, it results with the - flag specified and positive field width. (Note: This is the minimum width: The value is never truncated.)

. followed by integer number or *, or neither that specifies precision of the conversion. In the case when * is used, the precision is specified by an additional argument of type int. If the value of this argument is negative, it is ignored. If neither a number nor * is used, the precision is taken as zero. See the table below for exact effects of precision.

So if we try both conversion specification

#include <stdio.h>  int main() {     int precision = 8;     int biggerPrecision = 16;     const char *greetings = "Hello world";      printf("|%.8s|\n", greetings);     printf("|%.*s|\n", precision , greetings);     printf("|%16s|\n", greetings);     printf("|%*s|\n", biggerPrecision , greetings);      return 0; } 

we get the output:

|Hello wo| |Hello wo| |     Hello world| |     Hello world|