CODEWITHSUNDEEP
csyntax
Alex Walczak
Alex Walczak

Reputation: 1396

In C, does difference between char *s and char s[] apply to other types?

I have seen (1, 2) that:

char s[] = "abc";
char t[3] = "abc";

are effectively identical to:

char s[] = { 'a', 'b', 'c', '\0' };
char t[] = { 'a', 'b', 'c' };

And the following:

char *word = "abc";
word[0] = 'd';

places word in read-only memory, causing the illegal memory operation word[0] = 'd' to error.

Is this only the case with chars? I can't get an error when I do something like this:

int array[] = {1, 2, 3};
int *p = array;
p[0] = 0; // No error here
array[1] = 1; // or here

Upvotes: 3

Views: 382

Answers (2)

AnT stands with Russia
AnT stands with Russia

Reputation: 320361

There's no exact analog for string literals for other array types. But you can come somewhat close by using a const-qualified compound literal

const int *p = (const int []) { 1, 2, 3 };

There is a chance it will be placed into read-only memory, and might crash the program if an attempt is made to modify it (after casting away constness that is). Another similarity with string literals is that const-qualified compound literals might share storage.

One obvious difference between compound literals and string literals is that string literals always have static storage duration, while block-level compound literals have automatic storage duration.

Upvotes: 1

Keith Thompson
Keith Thompson

Reputation: 263167

It applies only to string literals.

A string literal implicitly creates an array object of type char[N] containing the character in the literal followed by a terminating \0' character. This object has static storage duration and is read-only. (It's not const, for historical reasons, but attempting to modify it has undefined behavior.)

When you write:

char *ptr = "abc";

you're creating ptr as a pointer object, and initializing it to point to the read-only static array containing "abc". (You can prevent attempts to modify it by defining it as const.)

When you write:

char arr[] = "abc";

you're creating arr as an array object, of type char[4], and copying the contents of the static read-only array into that object. arr is not read-only.

int array[] = {1, 2, 3};

creates an array object and initializes it as shown. There are no "integer array literals" that act like string literals. (There almost are -- see "compound literals" -- but those don't have the same read-only semantics as string literals.)

Note that this:

char t[3] = "abc";

is a special case: if an array is initialized with a string literal, and there isn't room for the terminating '\0', then only the non-null characters are copied to the array. As a result, t does not contain a string, just an unterminated sequence of characters. This isn't particularly relevant to your question.

Upvotes: 7

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