C strncpy 2 char

Question

I'm unable to make strncpy when I want to copy 2 chars from string, but same code is working when I want to copy 3 chars.

In this example, I need strncpy() to store 12 in 'to' variable:

void main(){
    const char* from = "12345678";
    char *to = (char*) malloc(3);
    strncpy(to, from, 2);

    printf("%s", to);
    free(to);
}

but when I use strncpy(to, from, 3) with malloc(4), it works ok. Any solution to make this work?

It is difficult to offer solutions when the problem statement is simply, "it doesn't work". Please [edit] your question to give a more complete description of what you expected to happen and how that differs from the actual results. See [ask] for hints on what makes a good explanation. — Toby Speight, Jun 30 '17 at 12:30
Just read the doc: No null-character is implicitly appended at the end of destination if source is longer than num. — Boiethios, Jun 30 '17 at 12:31
Hint for the future: Don't use `strncpy`, most of the time it does not do what you want. — Jabberwocky, Jun 30 '17 at 12:41
This bug is caused by using the dangerous `strncpy` function indeed. Never use that function, use `strcpy` or `memcpy` instead (together with appropriate buffer size checks). — Lundin, Jun 30 '17 at 14:02

score 3 · Answer 1 · answered Jun 30 '17 at 12:31

3

You forgot to append the destination string with a terminating zero.

char *to = (char*) malloc(3);
strncpy(to, from, 2);
to[2] = '\0';

answered Jun 30 '17 at 12:31

Vlad from Moscow

it works like that, thanks well if I use malloc (3) and need 2 chars wont \n be added at the end of that array? – ggoran Jun 30 '17 at 12:35
1

You do not need the cast in C – Ed Heal Jun 30 '17 at 12:35

1 Answers1