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cstringinitializationliterals
Dan Hulme
Dan Hulme

Reputation: 15290

Does a string literal count as a partial initializer and zero-initialize?

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.

Upvotes: 15

Views: 1418

Answers (3)

AnT stands with Russia
AnT stands with Russia

Reputation: 320671

Absolutely everywhere in C language it follows the all-or-nothing approach to initialization. If you initialize an aggregate only partially, the rest of that aggregate gets zero-initialized.

One can say that this is excessive and less than optimal with strings, but that's just how it works in C.

Upvotes: 3

hmjd
hmjd

Reputation: 122001

From the C99 standard (as already stated by ouah):

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.

and:

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static storage duration is not initialized explicitly, then:

  • if it has pointer type, it is initialized to a null pointer;
  • if it has arithmetic type, it is initialized to (positive or unsigned) zero;
  • if it is an aggregate, every member is initialized (recursively) according to these rules;
  • if it is a union, the first named member is initialized (recursively) according to these rules.

And char is an arithmetic type, so the remaining elements of the array will be initialised to zero.

Upvotes: 8

ouah
ouah

Reputation: 145899

 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."

Upvotes: 15

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