char *str = "string";

分配一个指向字符串字面量的指针，编译器将其放入可执行文件中不可修改的部分;

char str[] = "string";

分配并初始化一个可修改的本地数组

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).

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。

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.

不变的记忆

由于字符串字面量在设计上是只读的，所以它们存储在内存的Constant部分。存储在那里的数据是不可变的，即不能被更改。因此，在C代码中定义的所有字符串字面值在这里都获得一个只读内存地址。

栈内存

内存的堆栈部分是存放局部变量地址的地方，例如，函数中定义的变量。

正如@matli的回答所暗示的，有两种方法来处理这些常量字符串。

1. 指向字符串字面量的指针

当我们定义指向字符串字面量的指针时，我们是在Stack内存中创建一个指针变量。它指向底层字符串字面值所在的只读地址。

#include <stdio.h>

int main(void) {
  char *s = "hello";
  printf("%p\n", &s);  // Prints a read-only address, e.g. 0x7ffc8e224620
  return 0;
}

如果我们试图通过插入来修改s

s[0] = 'H';

我们得到一个分割错误(核心转储)。我们试图访问不应该访问的内存。我们正在尝试修改只读地址0x7ffc8e224620的值。

2. 字符数组

对于示例而言，假设存储在常量内存中的字符串字面值“Hello”具有与上述地址相同的只读内存地址0x7ffc8e224620。

#include <stdio.h>

int main(void) {
  // We create an array from a string literal with address 0x7ffc8e224620.
  // C initializes an array variable in the stack, let's give it address
  // 0x7ffc7a9a9db2.
  // C then copies the read-only value from 0x7ffc8e224620 into 
  // 0x7ffc7a9a9db2 to give us a local copy we can mutate.
  char a[] = "hello";

  // We can now mutate the local copy
  a[0] = 'H';

  printf("%p\n", &a);  // Prints the Stack address, e.g. 0x7ffc7a9a9db2
  printf("%s\n", a);   // Prints "Hello"

  return 0;
}

注意:当使用指针指向字符串字面量时，如1。，最好的做法是使用const关键字，如const *s = "hello"。这样可读性更强，并且当它被违反时，编译器将提供更好的帮助。然后它将抛出类似error:分配只读位置' *s '的错误，而不是seg错误。编辑器中的linter也可能在手动编译代码之前发现错误。

char *str = "string";

分配一个指向字符串字面量的指针，编译器将其放入可执行文件中不可修改的部分;

char str[] = "string";

分配并初始化一个可修改的本地数组