What is the difference between char array and char pointer in C?

Question

I am trying to understand pointers in C but I am currently confused with the following:

• char *p = \"hello\"
  

  This is a char pointer

Answer 1

What is the difference between char array vs char pointer in C?

C99 N1256 draft

There are two different uses of character string literals:

1. Initialize char[]:

   char c[] = "abc";      
   

   This is "more magic", and described at 6.7.8/14 "Initialization":

   An array of character type may be initialized by a character string literal, optionally enclosed in braces. Successive characters of the character string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.

   So this is just a shortcut for:

   char c[] = {'a', 'b', 'c', '\0'};
   

   Like any other regular array, c can be modified.

2. Everywhere else: it generates an:

   • unnamed
   • array of char What is the type of string literals in C and C++?
   • with static storage
   • that gives UB (undefined behavior) if modified

   So when you write:

   char *c = "abc";
   

   This is similar to:

   /* __unnamed is magic because modifying it gives UB. */
   static char __unnamed[] = "abc";
   char *c = __unnamed;
   

   Note the implicit cast from char[] to char *, which is always legal.

   Then if you modify c[0], you also modify __unnamed, which is UB.

   This is documented at 6.4.5 "String literals":

   5 In translation phase 7, a byte or code of value zero is appended to each multibyte character sequence that results from a string literal or literals. The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence. For character string literals, the array elements have type char, and are initialized with the individual bytes of the multibyte character sequence [...]

   6 It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.

6.7.8/32 "Initialization" gives a direct example:

EXAMPLE 8: The declaration

char s[] = "abc", t[3] = "abc";

defines "plain" char array objects s and t whose elements are initialized with character string literals.

This declaration is identical to

char s[] = { 'a', 'b', 'c', '\0' },
t[] = { 'a', 'b', 'c' };

The contents of the arrays are modifiable. On the other hand, the declaration

char *p = "abc";

defines p with type "pointer to char" and initializes it to point to an object with type "array of char" with length 4 whose elements are initialized with a character string literal. If an attempt is made to use p to modify the contents of the array, the behavior is undefined.

GCC 4.8 x86-64 ELF implementation

Program:

#include 

int main(void) {
    char *s = "abc";
    printf("%s\n", s);
    return 0;
}

Compile and decompile:

gcc -ggdb -std=c99 -c main.c
objdump -Sr main.o

Output contains:

 char *s = "abc";
8:  48 c7 45 f8 00 00 00    movq   $0x0,-0x8(%rbp)
f:  00 
        c: R_X86_64_32S .rodata

Conclusion: GCC stores char* it in .rodata section, not in .text.

If we do the same for char[]:

 char s[] = "abc";

we obtain:

17:   c7 45 f0 61 62 63 00    movl   $0x636261,-0x10(%rbp)

so it gets stored in the stack (relative to %rbp).

Note however that the default linker script puts .rodata and .text in the same segment, which has execute but no write permission. This can be observed with:

readelf -l a.out

which contains:

 Section to Segment mapping:
  Segment Sections...
   02     .text .rodata