programming - What is the difference between char s[] and char*s?



char* in c example (8)

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:

    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 <stdio.h>

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

https://ffff65535.com

In C, one can use a string literal in a declaration like this:

char s[] = "hello";

or like this:

char *s = "hello";

So what is the difference? I want to know what actually happens in terms of storage duration, both at compile and run time.


As an addition, consider that, as for read-only purposes the use of both is identical, you can access a char by indexing either with [] or *(<var> + <index>) format:

printf("%c", x[1]);     //Prints r

And:

printf("%c", *(x + 1)); //Prints r

Obviously, if you attempt to do

*(x + 1) = 'a';

You will probably get a Segmentation Fault, as you are trying to access read-only memory.


Given the declarations

char *s0 = "hello world";
char s1[] = "hello world";

assume the following hypothetical memory map:

                    0x01  0x02  0x03  0x04
        0x00008000: 'h'   'e'   'l'   'l'
        0x00008004: 'o'   ' '   'w'   'o'
        0x00008008: 'r'   'l'   'd'   0x00
        ...
s0:     0x00010000: 0x00  0x00  0x80  0x00
s1:     0x00010004: 'h'   'e'   'l'   'l'
        0x00010008: 'o'   ' '   'w'   'o'
        0x0001000C: 'r'   'l'   'd'   0x00

The string literal "hello world" is a 12-element array of char (const char in C++) with static storage duration, meaning that the memory for it is allocated when the program starts up and remains allocated until the program terminates. Attempting to modify the contents of a string literal invokes undefined behavior.

The line

char *s0 = "hello world";

defines s0 as a pointer to char with auto storage duration (meaning the variable s0 only exists for the scope in which it is declared) and copies the address of the string literal (0x00008000 in this example) to it. Note that since s0 points to a string literal, it should not be used as an argument to any function that would try to modify it (e.g., strtok(), strcat(), strcpy(), etc.).

The line

char s1[] = "hello world";

defines s1 as a 12-element array of char (length is taken from the string literal) with auto storage duration and copies the contents of the literal to the array. As you can see from the memory map, we have two copies of the string "hello world"; the difference is that you can modify the string contained in s1.

s0 and s1 are interchangeable in most contexts; here are the exceptions:

sizeof s0 == sizeof (char*)
sizeof s1 == 12

type of &s0 == char **
type of &s1 == char (*)[12] // pointer to a 12-element array of char

You can reassign the variable s0 to point to a different string literal or to another variable. You cannot reassign the variable s1 to point to a different array.


In the case of:

char *x = "fred";

x is an lvalue -- it can be assigned to. But in the case of:

char x[] = "fred";

x is not an lvalue, it is an rvalue -- you cannot assign to it.


Just to add: you also get different values for their sizes.

printf("sizeof s[] = %zu\n", sizeof(s));  //6
printf("sizeof *s  = %zu\n", sizeof(s));  //4 or 8

As mentioned above, for an array '\0' will be allocated as the final element.


The difference here is that

char *s = "Hello world";

will place "Hello world" in the read-only parts of the memory, and making s a pointer to that makes any writing operation on this memory illegal.

While doing:

char s[] = "Hello world";

puts the literal string in read-only memory and copies the string to newly allocated memory on the stack. Thus making

s[0] = 'J';

legal.


char *str = "Hello";

The above sets str to point to the literal value "Hello" which is hard-coded in the program's binary image, which is flagged as read-only in memory, means any change in this String literal is illegal and that would throw segmentation faults.

char str[] = "Hello";

copies the string to newly allocated memory on the stack. Thus making any change in it is allowed and legal.

means str[0] = 'M';

will change the str to "Mello".

For more details, please go through the similar question:

Why do I get a segmentation fault when writing to a string initialized with "char *s" but not "char s[]"?


char s[] = "Hello world";

Here, s is an array of characters, which can be overwritten if we wish.

char *s = "hello";

A string literal is used to create these character blocks somewhere in the memory which this pointer s is pointing to. We can here reassign the object it is pointing to by changing that, but as long as it points to a string literal the block of characters to which it points can't be changed.





constants