下面的代码在第2行接收到seg错误:
char *str = "string";
str[0] = 'z'; // could be also written as *str = 'z'
printf("%s\n", str);
虽然这个方法非常有效:
char str[] = "string";
str[0] = 'z';
printf("%s\n", str);
用MSVC和GCC测试。
当前回答
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.
其他回答
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。
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。
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。
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。
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.
