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?

Solution 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);
}

Solution 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|

Solution 3

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.

Solution 4

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

.* The precision is not specified in the format string, but as an additional integer value argument preceding the argument that has to be formatted.

s String of characters

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Comments

[Andrew Marshall]

Without additional arguments, that is not a valid printf call.

[Jonathan Leffler]

I've added the URL cross-reference so you can avoid charges of plagiarism. Of course, the correct quote says "The precision is not …" rather than "The width is not…".

[M.M]

It should be noted that the str_len argument must have type int (or narrower integral type, which would be promoted to int). It would be a bug to pass long, size_t, etc.

[Anton Samsonov]

As @MattMcNabb pointed out, every reference to that page must highlight that “an integer value” is exactly int (or a subset of it) — not just any integral value like more intuitive size_t or its possible aliases, like std::string::size_type. This is even more confusing, taking into account that the referenced page mentions size_t as one of supported type specifiers.

[Sonic Atom]

It's worth mentioning that the likely purpose of this code, especially when used with %s, is to print a sub-string of the original string. In this use case, str would point to somewhere inside the original string (possibly at the beginning), and str_len will specify the length of the sub-string that should be printed.

[Admin]

By specifying a length, we can get around printing (or sprintf) 'ing a string which has no null terminator, for example a string which is input from any stream or file based source. Which is far more often the use case I have encountered, than merely print prettines.

[powersource97]

From the docs, the major difference between '%.*s' and '%*s' seems to be that the former ignore negative values for precision while the later takes in into consideration to apply the appropriate justification.

[kenn]

You could also add printf("|%.*s|\n", biggerPrecision , greetings);