要理解这个错误或问题，您应该首先了解指针和数组的差异b/w 所以在这里，我首先要解释一下它们的区别

字符串数组

 char strarray[] = "hello";

在存储器数组中存储的是连续存储器单元，存储为[h][e][l][l][o][\0] =>[]是1个char字节大小的存储器单元，而这个连续存储器单元可以通过名为strarray的名称在这里访问。这里string数组strarray本身包含初始化的所有字符串。在这种情况下，"hello" 因此，我们可以通过访问每个字符的索引值来轻松地更改其内存内容

`strarray[0]='m'` it access character at index 0 which is 'h'in strarray

它的值变成了m所以strarray的值变成了mello;

这里需要注意的一点是，我们可以通过一个字符一个字符地改变字符串数组的内容，但不能像strarray="new string"这样直接初始化其他字符串，这是无效的

指针

我们都知道指针指向内存中的内存位置， 未初始化的指针指向随机内存位置，初始化后指向特定内存位置

char *ptr = "hello";

这里的指针ptr被初始化为字符串“hello”，这是一个存储在只读存储器(ROM)中的常量字符串，所以“hello”不能被更改，因为它存储在ROM中

PTR存储在堆栈部分并指向常量字符串"hello"

所以ptr[0]='m'是无效的，因为你不能访问只读内存

但是ptr可以直接初始化为其他字符串值，因为它只是一个指针，所以它可以指向其数据类型变量的任何内存地址

ptr="new string"; is valid

当您试图访问不可访问的内存时，会导致分割错误。

Char *str是一个指向不可修改的字符串的指针(这是导致segfault的原因)。

而char str[]是一个数组，可以修改。

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.

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。

// 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. 

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.