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

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.

其他回答

// 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. 

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。

因为在第一个例子的上下文中,“whatever”的类型是const char*(即使你将它赋值给一个非const char*),这意味着你不应该尝试写它。

编译器通过将字符串放在内存的只读部分来强制执行这一点,因此写入它会产生段错误。

首先,str是一个指向"string"的指针。编译器允许将字符串字面量放在内存中不能写入,但只能读取的地方。(这真的应该触发一个警告,因为你将一个const char *分配给一个char *。你是禁用了警告,还是忽略了它们?)

第二,你在创建一个数组,它是你可以完全访问的内存,并用"string"初始化它。您正在创建一个字符[7](六个用于字母,一个用于结尾的'\0'),您可以对它做任何您喜欢的事情。

The

 char *str = "string";

Line定义了一个指针,并将其指向一个字面值字符串。字面值字符串是不可写的,所以当你这样做:

  str[0] = 'z';

你会得到一个隔离失误。在某些平台上,字面值可能位于可写内存中,因此您不会看到段错误,但无论如何它都是无效代码(导致未定义的行为)。

线:

char str[] = "string";

分配一个字符数组并将字面值字符串复制到该数组中,该数组是完全可写的,因此后续更新没有问题。