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

Solution 1:

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.

Solution 2:

First off, in function arguments, they are exactly equivalent:

void foo(char *x);
void foo(char x[]); // exactly the same in all respects

In other contexts, char * allocates a pointer, while char [] allocates an array. Where does the string go in the former case, you ask? The compiler secretly allocates a static anonymous array to hold the string literal. So:

char *x = "Foo";
// is approximately equivalent to:
static const char __secret_anonymous_array[] = "Foo";
char *x = (char *) __secret_anonymous_array;

Note that you must not ever attempt to modify the contents of this anonymous array via this pointer; the effects are undefined (often meaning a crash):

x[1] = 'O'; // BAD. DON'T DO THIS.

Using the array syntax directly allocates it into new memory. Thus modification is safe:

char x[] = "Foo";
x[1] = 'O'; // No problem.

However the array only lives as long as its contaning scope, so if you do this in a function, don't return or leak a pointer to this array - make a copy instead with strdup() or similar. If the array is allocated in global scope, of course, no problem.

Solution 3:

This declaration:

char s[] = "hello";

Creates one object - a char array of size 6, called s, initialised with the values 'h', 'e', 'l', 'l', 'o', '\0'. Where this array is allocated in memory, and how long it lives for, depends on where the declaration appears. If the declaration is within a function, it will live until the end of the block that it is declared in, and almost certainly be allocated on the stack; if it's outside a function, it will probably be stored within an "initialised data segment" that is loaded from the executable file into writeable memory when the program is run.

On the other hand, this declaration:

char *s ="hello";

Creates two objects:

• a read-only array of 6 chars containing the values 'h', 'e', 'l', 'l', 'o', '\0', which has no name and has static storage duration (meaning that it lives for the entire life of the program); and
• a variable of type pointer-to-char, called s, which is initialised with the location of the first character in that unnamed, read-only array.

The unnamed read-only array is typically located in the "text" segment of the program, which means it is loaded from disk into read-only memory, along with the code itself. The location of the s pointer variable in memory depends on where the declaration appears (just like in the first example).

Solution 4:

Given the declarations

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

assume the following hypothetical memory map (the columns represent characters at offsets 0 to 3 from the given row address, so e.g. the 0x00 in the bottom right corner is at address 0x0001000C + 3 = 0x0001000F):

                     +0    +1    +2    +3
        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.