C: differences between char pointer and array [dup

Question

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• Why do I get a segmentation fault when writing to a string initialized with “char *s” but not “char s[]”? 16 answers

Consider:

char amessage[] = "now is the time";
char *pmessage = "now is the time";

I read from The C Programming Language, 2nd Edition that the above two statements don't do the same thing.

I always thought that an array is an convenient way to manipulate pointers to store some data, but this is clearly not the case... What are the "non-trivial" differences between arrays and pointers in C?

Answer 1

I can't add usefully to the other answers, but I will remark that in Deep C Secrets, Peter van der Linden covers this example in detail. If you are asking these kinds of questions I think you will love this book.

---

P.S. You can assign a new value to pmessage. You can't assign a new value to amessage; it is immutable.

Answer 2

The first form (amessage) defines a variable (an array) which contains a copy of the string "now is the time".

The second form (pmessage) defines a variable (a pointer) that lives in a different location than any copy of the string "now is the time".

Try this program out:

#include <inttypes.h>
#include <stdio.h>

int main (int argc, char *argv [])
{
     char  amessage [] = "now is the time";
     char *pmessage    = "now is the time";

     printf("&amessage   : %#016"PRIxPTR"\n", (uintptr_t)&amessage);
     printf("&amessage[0]: %#016"PRIxPTR"\n", (uintptr_t)&amessage[0]);
     printf("&pmessage   : %#016"PRIxPTR"\n", (uintptr_t)&pmessage);
     printf("&pmessage[0]: %#016"PRIxPTR"\n", (uintptr_t)&pmessage[0]);

     printf("&\"now is the time\": %#016"PRIxPTR"\n",
            (uintptr_t)&"now is the time");

     return 0;
}

You'll see that while &amessage is equal to &amessage[0], this is not true for &pmessage and &pmessage[0]. In fact, you'll see that the string stored in amessage lives on the stack, while the string pointed at by pmessage lives elsewhere.

The last printf shows the address of the string literal. If your compiler does "string pooling" then there will be only one copy of the string "now is the time" -- and you'll see that its address is not the same as the address of amessage. This is because amessage gets a copy of the string when it is initialized.

In the end, the point is that amessage stores the string in its own memory (on the stack, in this example), while pmessage points to the string which is stored elsewhere.

Answer 3

The second one allocates the string in some read-only section of the ELF. Try the following:

#include <stdio.h>

int main(char argc, char** argv) {
    char amessage[] = "now is the time";
    char *pmessage = "now is the time";

    amessage[3] = 'S';
    printf("%s\n",amessage);

    pmessage[3] = 'S';
    printf("%s\n",pmessage);
}

and you will get a segfault on the second assignment (pmessage[3]='S').

Answer 4

True, but it's a subtle difference. Essentially, the former:

char amessage[] = "now is the time";

Defines an array whose members live in the current scope's stack space, whereas:

char *pmessage = "now is the time";

Defines a pointer that lives in the current scope's stack space, but that references memory elsewhere (in this one, "now is the time" is stored elsewhere in memory, commonly a string table).

Also, note that because the data belonging to the second definition (the explicit pointer) is not stored in the current scope's stack space, it is unspecified exactly where it will be stored and should not be modified.

Edit: As pointed out by Mark, GMan, and Pavel, there is also a difference when the address-of operator is used on either of these variables. For instance, &pmessage returns a pointer of type char**, or a pointer to a pointer to chars, whereas &amessage returns a pointer of type char(*)[16], or a pointer to an array of 16 chars (which, like a char**, needs to be dereferenced twice as litb points out).

Answer 5

Here's a hypothetical memory map, showing the results of the two declarations:

                0x00  0x01  0x02  0x03  0x04  0x05  0x06  0x07
    0x00008000:  'n'   'o'   'w'   ' '   'i'   's'   ' '   't'
    0x00008008:  'h'   'e'   ' '   't'   'i'   'm'   'e'  '\0'
        ...
amessage:
    0x00500000:  'n'   'o'   'w'   ' '   'i'   's'   ' '   't'
    0x00500008:  'h'   'e'   ' '   't'   'i'   'm'   'e'  '\0'
pmessage:
    0x00500010:  0x00  0x00  0x80  0x00

The string literal "now is the time" is stored as a 16-element array of char at memory address 0x00008000. This memory may not be writable; it's best to assume that it's not. You should never attempt to modify the contents of a string literal.

The declaration

char amessage[] = "now is the time";

allocates a 16-element array of char at memory address 0x00500000 and copies the contents of the string literal to it. This memory is writable; you can change the contents of amessage to your heart's content:

strcpy(amessage, "the time is now");

The declaration

char *pmessage = "now is the time";

allocates a single pointer to char at memory address 0x00500010 and copies the address of the string literal to it.

Since pmessage points to the string literal, it should not be used as an argument to functions that need to modify the string contents:

strcpy(amessage, pmessage); /* OKAY */
strcpy(pmessage, amessage); /* NOT OKAY */
strtok(amessage, " ");      /* OKAY */
strtok(pmessage, " ");      /* NOT OKAY */
scanf("%15s", amessage);      /* OKAY */
scanf("%15s", pmessage);      /* NOT OKAY */

and so on. If you changed pmessage to point to amessage:

pmessage = amessage;

then it can be used everywhere amessage can be used.

Answer 6

A pointer is just a variable that holds a memory address. Notice that you are playinf with "string literals" which is another issue. Differences explained inline: Basically:

#include <stdio.h>

int main ()
{

char amessage[] = "now is the time"; /* Attention you have created a "string literal" */

char *pmessage = "now is the time";  /* You are REUSING the string literal */


/* About arrays and pointers */

pmessage = NULL; /* All right */
amessage = NULL; /* Compilation ERROR!! */

printf ("%d\n", sizeof (amessage)); /* Size of the string literal*/
printf ("%d\n", sizeof (pmessage)); /* Size of pmessage is platform dependent - size of memory bus (1,2,4,8 bytes)*/

printf ("%p, %p\n", pmessage, &pmessage);  /* These values are different !! */
printf ("%p, %p\n", amessage, &amessage);  /* These values are THE SAME!!. There is no sense in retrieving "&amessage" */


/* About string literals */

if (pmessage == amessage)
{
   printf ("A string literal is defined only once. You are sharing space");

   /* Demostration */
   "now is the time"[0] = 'W';
   printf ("You have modified both!! %s == %s \n", amessage, pmessage);
}


/* Hope it was useful*/
return 0;
}