C：printf char *返回奇怪的最后一个char

Question

I'm trying to understand why I've got a weird character after my printf()

char* extract_word()
{
    char* sentences = "hi! i'm a banana!";
    int starts = 4;
    int ends   = 12;
    int count;
    int nb_char = ends-starts+1;
    char* word = malloc(nb_char);
    printf("\n\n%d\n",ends-starts);
    for(count = starts; count < ends;count++)
    {
        word[count-starts] = sentences[count];
        printf("%c == \n",sentences[count]);
    }
    word[count-starts+1] = '\0';
    printf("\n\n%s",word);
    return word;
}

The printf returns:

 8 i == ' == m == == a == == b == a == i'm a bau 

If I remove the '\\0' I get something like:

  'ma ba¨Á£´ 

Answer 1

In your code

   word[count-starts+1] = '\0';

is off-by-one and basically that out-of-bound access invokes undefined behavior .

You should change your code to

   word[nb_char-1] = '\0';

because, you have allocated nb_char bytes and the last index would be nb_char-1 .

That said, it's always required to check for the success of malloc() by checking the return against NULL before using the returned pointer.

Answer 2

If you remove the \\0 printf has no way to know that the strings ends and will keep incrementing the pointer until it sees a null value and you take the risks of getting a segmentation fault.

For strings without a 0 at the end you can use snprintf

Also for what you are trying to achieve there's memcpy or strncpy

check the man pages for more details.