memory for array initialization with string literal

Question

I was going through this QA where it is said that char array when initialized with string literal will cause two memory allocations one for variable and other for string literal.

I have written below program to see how is the memory allocated.

#include <stdio.h>
#include <string.h>

int main()
{
    char a[] = "123454321";
    
    printf("a =%p and &a = %p\n", a, &a);

    for(int i = 0; i< strlen(a); i++)
        printf("&a[%d] =%p and a[%d] = %c\n",i,&a[i],i,a[i]);
    
    return 0;
}

and the output is:

a =0x7ffdae87858e and &a = 0x7ffdae87858e                                                                             
&a[0] =0x7ffdae87858e and a[0] = 1                                                                                    
&a[1] =0x7ffdae87858f and a[1] = 2                                                                                    
&a[2] =0x7ffdae878590 and a[2] = 3                                                                                    
&a[3] =0x7ffdae878591 and a[3] = 4                                                                                    
&a[4] =0x7ffdae878592 and a[4] = 5                                                                                    
&a[5] =0x7ffdae878593 and a[5] = 4                                                                                    
&a[6] =0x7ffdae878594 and a[6] = 3                                                                                    
&a[7] =0x7ffdae878595 and a[7] = 2                                                                                    
&a[8] =0x7ffdae878596 and a[8] = 1

From the output it does not look like we have two separate memory locations for array and string literal.

If we have separate memory for array and string literal, is there any way we can prove array a and string literal stores separately in this scenario?

link to clone: https://onlinegdb.com/HkJhdSHyd

Answer 1

char a[] = "123454321";

Technically, the string literal "123454321" is not required to be stored anywhere as such. All that's required is that a[] be initialized with the right values when main is entered. Whether that's done by copying the string from some static read-only memory location, or running code that fills it in some other way is not mandated by the standard.

As far as the standard goes, it would be perfectly acceptable for the compiler to emit code equivalent to the following in order to initialize a[]:

char a[10];
for(int n = 0; n <= 4; n++)
    a[n] = a[8-n] = '1' + n;
a[9] = '\0';

In fact, at least one compiler (gcc) initializes a[] via custom code, rather than storing and copying the literal string.

mov     DWORD PTR [ebp-22], 875770417    ; =  0x34333231  =  '1', '2', '3', '4'
mov     DWORD PTR [ebp-18], 842216501    ; =  0x32333435  =  '5`, '4', '3', '2'
mov     WORD  PTR [ebp-14], 49           ; =  0x31        =  '1', '\0'

Answer 2

You've completely misunderstood the question and answer. The question was about whether the initializer string consumes memory in addition to the actual array. Now the thing is, you cannot observe the initializer string.

It is like there are two sheets of paper. One in the closet with 123454321 written with ballpoint pen. One on the desk - initially empty. Then someone else comes, takes the sheet from the closet, reads the text on it, and writes it on the sheet on the desk using a pencil. Then puts the paper back into closet.

Now you're looking at that sheet on desk saying: "clearly the text 123454321 has not been written twice onto this sheet, hence what do they say about there being two copies?"

Answer 3

You can prove it by modifying the code as follows:

int main()
{
    for (int i = 0; i < 2; ++i)
    {
        char a[] = "123454321";

        printf("a = %s\n", a);
        a[3] = 'x';
        a[5] = 'y';
        printf("a = %s\n", a);
    }
}

Output:

a = 123454321
a = 123x5y321
a = 123454321
a = 123x5y321

We got the original string back after modifying it, so the original string must have been stored somewhere other than the place we modified.

Answer 4

You cant prove there's two storages because you have only one.

The compiler sees you want a char array initialized with some characters and '\0' so it does that. It does not need to store the string literal somewhere else.

This would not compile for that reason.

#include <stdio.h>
#include <string.h>

char *p = "123454321";

int main()
{
    char a[] = p;
    
    printf("a =%p and &a = %p\n", a, &a);

    for(int i = 0; i< strlen(a); i++)
        printf("&a[%d] =%p and a[%d] = %c\n",i,&a[i],i,a[i]);
    
    return 0;
}