How do I print a non-null-terminated string using printf?

How can I print a non-null-terminated string using printf, assuming that I know the length of the string at runtime?

printf("%.*s", length, string);

Use together with other args:

printf("integer=%d, string=%.*s, number=%f", integer, length, string, number);
//                         ^^^^                       ^^^^^^^^^^^^^^

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In C you could specify the maximum length to output with the %.123s format. This means the output length is at most 123 chars. The 123 could be replaced by *, so that the length will be taken from the argument of printf instead of hard-coded.

Note that this assumes the string does not contain any interior null bytes (\0), as %.123s only constrains the maximum length not the exact length, and strings are still treated as null-terminated.

If you want to print a non-null-terminated string with interior null, you cannot use a single printf. Use fwrite instead:

fwrite(string, 1, length, stdout);

See @M.S.Dousti's answer for detailed explanation.

The answer provided by @KennyTM is great, but with a subtlety.

In general, if the string is non-null "terminated", but has a null character in the middle, printf("%.*s", length, string); does not work as expected. This is because the %.*s format string asks printf to print a maximum of length characters, not exactly length characters.

I'd rather use the more general solution pointed out by @William Pursell in a comment under the OP:

fwrite(string, sizeof(char), length, stdout);

Here's a sample code:

#include <stdio.h>

int main(void) {
    size_t length = 5;

    char string[length];
    string[0] = 'A';
    string[1] = 'B';
    string[2] = 0;        // null character in the middle
    string[3] = 'C';
    string[4] = 'D';

    printf("With printf: %.*s\n", length, string);
    printf("With fwrite: ");
    fwrite(string, sizeof(char), length, stdout);
    printf("\n");

    return 0;
}

Output:

With printf: AB
With fwrite: AB CD