Thread: Memory Layout of character pointers

Because you are printing the value and not the address.

names[i] // Value

&names[i] // Address

Originally Posted by stahta01

Because you are printing the value and not the address.

names[i] // Value
&names[i] // Address

Nope... in this case 'names' is converted to a pointer and each element of the array is a pointer too...
To print a pointer, use %p, not %u.

More: There is no garantees where the compier will allocate the blocks of data, except for the individual string literals and the elements of array 'names' itself... In the example:

Code:

char *names[] = {  "t", "th", "str" };

The blocks 't\0', 'th\0' and 'str\0' could be allocated out of sequence. In assembly, the compiler could choose to do something like this:

Code:

section .data
str.names0:  db  't',0
i:           dd 0
str.names1:  db  't','h',0
names:       dq  str.names0, str.names1, str.names2
str.names2:  db  's', 't', 'r', 0

Except for the fact that 'str.names0', 'str.names1' and str.names2' are allocated in a different segment (.rodata), but, disconsidering this detail, the choice, above is still valid.

PS: Ok... i (if no otimizations were enabled) and names would be allocated on stack, but you get my point, don't ya?

Originally Posted by karthigan

Thank you for the detailed explanation! I used %p to print the pointer. And in this case, it is indeed continuous allocation - see output below:

Base Address of names : 0x561e9e1737d8
Base Address of string:0: 0x561e9e1737d8
Base Address of string:1: 0x561e9e1737da
Base Address of string:2: 0x561e9e1737dd

You're welcome.... But keep this in mind... it is not garanteed you got an sequential allocation, except for the array and individual string literals (which are pointers to arrays as well).

If you need that kind of garantee you could do:

Code:

static char *s = "t\0th\0str";
char *names[] = { s, s + 2, s + 5 };

And you are, still, doing something wrong... 'names' is allocated on stack, and its elements on .rodata section. They have different addresses:

Code:

#include <stdio.h>

int main( void )
{
  int i;
  char *names[] = { "t", "th", "str" };

  printf( "%p\n", names );

  for ( i = 0; i < 3; i++)
    printf( "%p\n", names[i] );
}

Compiling and running (x86-64):

Code:

$ cc -O2 -o test test.c
$ ./test
0x7ffc1c0d7620
0x55ff8695b804
0x55ff8695b806
0x55ff8695b809

Notice the different address form of 'names' (the first value).

Maybe drawing makes easier to see why 'name' and each element from the 'name' array cannot share their addresses:



I'm not having fun at your expense here... drawing really helps a lot!