Question

Normally, if you do the following:

int * i = &someint;

It’s just a pointer to a variable.

But, when you do

char * str = "somestring";

it automatically turns it into an array. Is it the pointer which is doing this, or is it just syntactic sugar for initialization syntax?

Answer 1

No, the string literal "somestring" is already a character array, almost certainly created by your compiler.

What that statement is doing is setting str to point to the first character. If you were to look at the underlying assembler code, it would probably look like:

str314159:  db   "somestring", 0  ; all string literals here.
: :         : :
            load r0, str314159    ; get address of string
            stor r0, -24[sp]      ; store it into local var str.

In a large number of cases, an array will decay to a pointer to the first element of that array (with some limited exceptions such as when doing sizeof).

---

By way of example, the following C code:

#include <stdio.h>

int main (void) {
    char *somestr = "Hello";
    puts (somestr);
    return 0;
}

when compiled with gcc -S to generate x86 assembly, gives us (with irrelevant cruft removed):

.LC0:
    .string    "Hello"
    .text
.globl main
    .type      main, @function
main:
    pushl      %ebp                ; Standard set up of stack frame,
    movl       %esp, %ebp          ;   aligning and making
    andl       $-16, %esp          ;   room for
    subl       $32, %esp           ;   local variables.

    movl       $.LC0, 28(%esp)     ; Load address of string in to somestr.

    movl       28(%esp), %eax      ; Call puts with that variable.
    movl       %eax, (%esp)
    call       puts

    movl       $0, %eax            ; Set return code.

    leave                          ; Tear down stack frame and return.
    ret

You can see that the address of the first character, .LC0, is indeed loaded into the somestr variable. And, while it may not be immediately obvious .string does create an array of characters terminated by the NUL character.