Question

In C, you can partially initialize a struct or array, with the result that the members/elements that aren’t mentioned in the initializer are zero-initialized. (C99 section 6.7.8.19). For example:-

int a[4] = {1, 2};
// a[0] == 1
// a[1] == 2
// a[2] == 0
// a[3] == 0

You can also initialize “an array of character type” with a string literal (C99 section 6.7.8.14), and “successive characters … initialize the elements of the array”. For example:-

char b[4] = "abc";
// b[0] == 'a'
// b[1] == 'b'
// b[2] == 'c'
// b[3] == '\0'

All pretty straightforward. But what happens if you explicitly give the length of the array, but use a literal that’s too short to fill the array? Are the remaining characters zero-initialized, or do they have undefined values?

char c[4] = "a";
// c[0] == 'a'
// c[1] == '\0'
// c[2] == ?
// c[3] == ?

Treating it as a partial initializer would make sense, it would make char c[4] = "a" behave exactly like char c[4] = {'a'}, and it would have the useful side-effect of letting you zero-initialize a whole character array concisely with char d[N] = "", but it’s not at all clear to me that that’s what the spec requires.

Answer 1

 char c[4] = "a";

All the remaining elements of the array will be set to 0. That is, not only c[1] but also c[2] and c[3].

Note that this does not depend on the storage duration of c, i. e., even if c has automatic storage duration the remaining elements will be set to 0.

From the C Standard (emphasis mine):

(C99, 6.7.8p21) “If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration.“