C如何处理本地字符串文字的内存？

Question

Consider this function:

void useless() {
   char data[] = "aaa";
}

From what I learned here , the "aaa" literal lives to the end of the program. However, the data[] (initialized by the literal) is local, so it lives only to the end of the function.

The memory is copied, so the program needs 4B for the literal, 4B for the data and sizeof(size_t) bytes for the pointer to data and sizeof(size_t) for the pointer of the literal - is this true?

If the literal has static storage duration, no new memory is allocated for the local literal by the second call - is this true?

Answer 1

   char data[] = "aaa";

This is not a string literal but just an array. So there's no pointer there and memory is allocated only for the data .

If the literal has static storage duration, no new memory is allocated for the local literal by the second call

This is true for string literals like: char *s="aaa"; From the standard:

2.13. Sttring literals
[...]An ordinary string literal has type “array of n const char” and static storage duration (3.7)

Answer 2

There is no pointer variable here. All there is an array, which is 4 bytes.

The compiler may or may not store the literal itself in memory; if it does, that is another 4 bytes.
Note that any memory taken up by anything other than the array itself is implementation-dependent.

I'm not sure what you mean by the "second call", but in general when you create an array, it takes up some amount of size... so if you create two arrays with the same literal, the compiler allocates space for two arrays (and perhaps -- or perhaps not -- for the literal also).