I got a code snippet in which there is a
printf("%.*s\n")
what does the %.*s mean?
4 Answers
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);
}
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|
I don't think the code above is correct but (according to this description of printf()) the .* means
The width is not specified in the format string, but as an additional integer value argument preceding the argument that has to be formatted.'
So it's a string with a passable width as an argument.
See:
.*The precision is not specified in the format string, but as an additional integer value argument preceding the argument that has to be formatted.
sString of characters
