C取消指向字符数组的指针

Question

I'm trying to grasp C pointers with this small example I found in a tutorial:

#include <stdio.h>

int main() {
    char vowels[] = {'A', 'E', 'I', 'O', 'U'};
    char *pvowels = &vowels;

    printf(pvowels);
//  printf(*pvowels);
    return 0;
}

If I compile this I get a warning "Initialization from incompatable pointer type" but it will still compile/run, printing AEIOU plus some random junk characters. I was also under the assumption that *pvowels would be the character at the first memory location of the vowels array and hence print an 'A' but it just segfaults instead.

My two questions are:

Why does the 1st printf print out the vowels + junk?

Why does the 2nd print not print out just an A?

Answer 1

When you use &vowels you get a pointer to the array not the first element in the array. The type of &vowels is char(*)[5] .

To fix this problem you should either use &vowels[0] to get a pointer to the first element, or rely on that arrays naturally decays to pointers to their first element:

char *pvowels = vowels;

The second problem is that you treat pvowels like a string , but a string is a sequence of characters with a terminator . Since the array vowels doesn't have this terminator the printf call will go out of bounds and you will have undefined behavior .

As for the last problem, when you dereference a pointer you get the value of the location where it points. Because the location of the array starts with the first element of the array both &vowels[0] and &vowels points to the same location, so when you dereference pvowels you get the first element of the array which is the character 'A' . However you can't use it as the first argument for printf because the printf function expects it to be a string (a pointer to a zero-terminated array of characters), not a single character.