参见C常见问题，问题1.32

Q: What is the difference between these initializations? char a[] = "string literal"; char *p = "string literal"; My program crashes if I try to assign a new value to p[i]. A: A string literal (the formal term for a double-quoted string in C source) can be used in two slightly different ways: As the initializer for an array of char, as in the declaration of char a[] , it specifies the initial values of the characters in that array (and, if necessary, its size). Anywhere else, it turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. In an expression context, the array is converted at once to a pointer, as usual (see section 6), so the second declaration initializes p to point to the unnamed array's first element. Some compilers have a switch controlling whether string literals are writable or not (for compiling old code), and some may have options to cause string literals to be formally treated as arrays of const char (for better error catching).

参见C常见问题，问题1.32

Q: What is the difference between these initializations? char a[] = "string literal"; char *p = "string literal"; My program crashes if I try to assign a new value to p[i]. A: A string literal (the formal term for a double-quoted string in C source) can be used in two slightly different ways: As the initializer for an array of char, as in the declaration of char a[] , it specifies the initial values of the characters in that array (and, if necessary, its size). Anywhere else, it turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. In an expression context, the array is converted at once to a pointer, as usual (see section 6), so the second declaration initializes p to point to the unnamed array's first element. Some compilers have a switch controlling whether string literals are writable or not (for compiling old code), and some may have options to cause string literals to be formally treated as arrays of const char (for better error catching).

// create a string constant like this - will be read only
char *str_p;
str_p = "String constant";

// create an array of characters like this 
char *arr_p;
char arr[] = "String in an array";
arr_p = &arr[0];

// now we try to change a character in the array first, this will work
*arr_p = 'E';

// lets try to change the first character of the string contant
*str_p = 'G'; // this will result in a segmentation fault. Comment it out to work.


/*-----------------------------------------------------------------------------
 *  String constants can't be modified. A segmentation fault is the result,
 *  because most operating systems will not allow a write
 *  operation on read only memory.
 *-----------------------------------------------------------------------------*/

//print both strings to see if they have changed
printf("%s\n", str_p); //print the string without a variable
printf("%s\n", arr_p); //print the string, which is in an array. 

首先是一个不能修改的常量字符串。第二个是一个初始化值的数组，因此它可以被修改。

假设字符串是，

char a[] = "string literal copied to stack";
char *p  = "string literal referenced by p";

在第一种情况下，当'a'进入作用域时，文字将被复制。这里'a'是定义在stack上的数组。这意味着字符串将在堆栈上创建，其数据从代码(文本)内存中复制，通常是只读的(这是特定于实现的，编译器也可以将这种只读的程序数据放在可读写内存中)。

在第二种情况下，p是定义在堆栈(本地作用域)上的指针，并引用存储在其他位置的字符串字面量(程序数据或文本)。通常，修改这样的内存不是好的实践，也不鼓励。

像“String”这样的字符串文字可能在可执行文件的地址空间中作为只读数据分配(通过编译器)。当你去触摸它时，它会害怕你在它的泳衣区，并让你知道一个隔离错误。

在第一个例子中，你得到一个指向const数据的指针。在第二个示例中，使用const数据的副本初始化一个7个字符的数组。