Why doesn't the compiler detect out-of-bounds in string constant initialization?

Question

I read this question and its answer in a book. But I didn\'t understand the book\'s justification.

Will the following code compile?

Answer 1

Your book must be pretty old, because gcc puts out a warning even without -Wall turned on:

$ gcc c.c
c.c: In function `main':
c.c:6: warning: initializer-string for array of chars is too long

If we slightly update the program:

#include 

int main(int argc, char **argv)
{

        char str[5] = "1234567890";
        printf("%s\n", str);
        return 0;
}

We can see that gcc seems to truncate the string to the length you've specified; I'm assuming that there happens to be a '\0' where str[6] would be, because otherwise we should see garbage after the 5; but maybe gcc implicitly makes str an array of length 6 and automatically sticks the '\0' in there - I'm not sure.

$ gcc c.c && ./a.exe
c.c: In function `main':
c.c:6: warning: initializer-string for array of chars is too long
12345