Are the braces around the string literal in char array declaration valid? (For example, char s [] = {& ldquo; Hello World & rdquo;})


By accident I found that the line char s[] = {"Hello World"}; is properly compiled and seems to be treated the same as char s[] = "Hello World";. Isn't the first ({"Hello World"}) an array containing one element that is an array of char, so the declaration for s should read char *s[]? In fact if I change it to char *s[] = {"Hello World"}; the compiler accepts it as well, as expected.

Searching for an answer, the only place I found which mentioned this is this one but there is no citing of the standard.

So my question is, why the line char s[] = {"Hello World"}; is compiled although the left side is of type array of char and the right side is of type array of array of char?

Following is a working program:

int main() {
    char s[] = {"Hello World"};
    printf("%s", s); // Same output if line above is char s[] = "Hello World";
    return 0;

Thanks for any clarifications.

P.S. My compiler is gcc-4.3.4.

It's allowed because the standard says so: C99 section 6.7.8, §14:

An array of character type may be initialized by a character string literal, optionally enclosed in braces. Successive characters of the character string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.

What this means is that both

char s[] = { "Hello World" };


char s[] = "Hello World";

are nothing more than syntactic sugar for

char s[] = { 'H', 'e', 'l', 'l', 'o', ' ', 'W', 'o', 'r', 'l', 'd', 0 };

On a related note (same section, §11), C also allows braces around scalar initializers like

int foo = { 42 };

which, incidentally, fits nicely with the syntax for compound literals

(int){ 42 }