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

问题:

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

回答1:

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

Use together with other args:

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

---

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.

回答2:

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   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 

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