通常，当程序运行时，字符串字面值存储在只读内存中。这是为了防止您意外地更改字符串常量。在第一个例子中，"string"存储在只读内存中，*str指向第一个字符。当您试图将第一个字符更改为'z'时，会发生段错误。

在第二个例子中，字符串"string"被编译器从其只读母数组复制到str[]数组中。然后允许更改第一个字符。你可以通过打印每个地址来检查:

printf("%p", str);

同样，在第二个例子中打印str的大小会显示编译器已经为它分配了7个字节:

printf("%d", sizeof(str));

char *str = "string";  

上面的代码将str设置为指向在程序的二进制映像中硬编码的字面值“string”，它在内存中可能被标记为只读。

因此str[0]=试图写入应用程序的只读代码。我猜这可能依赖于编译器。

在第一个代码中，"string"是一个字符串常量，字符串常量永远不应该被修改，因为它们通常被放置在只读内存中。"str"是一个用来修改常量的指针。

在第二段代码中，"string"是一个数组初始化器，类似于

char str[7] =  { 's', 't', 'r', 'i', 'n', 'g', '\0' };

"str"是堆栈上分配的数组，可以自由修改。

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。

The C FAQ that @matli linked to mentions it, but no one else here has yet, so for clarification: if a string literal (double-quoted string in your source) is used anywhere other than to initialize a character array (ie: @Mark's second example, which works correctly), that string is stored by the compiler in a special static string table, which is akin to creating a global static variable (read-only, of course) that is essentially anonymous (has no variable "name"). The read-only part is the important part, and is why the @Mark's first code example segfaults.

5.5节K&R的字符指针和功能也讨论了这个主题:

There is an important difference between these definitions: char amessage[] = "now is the time"; /* an array */ char *pmessage = "now is the time"; /* a pointer */ amessage is an array, just big enough to hold the sequence of characters and '\0' that initializes it. Individual characters within the array may be changed but amessage will always refer to the same storage. On the other hand, pmessage is a pointer, initialized to point to a string constant; the pointer may subsequently be modified to point elsewhere, but the result is undefined if you try to modify the string contents.