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。

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。

为什么我得到一个分割错误时写入字符串?

c99n1256草案

字符串字面量有两种不同的用法:

Initialize char[]: char c[] = "abc"; This is "more magic", and described at 6.7.8/14 "Initialization": An array of character type may be initialized by a character string literal, optionally enclosed in braces. Successive characters of the character string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array. So this is just a shortcut for: char c[] = {'a', 'b', 'c', '\0'}; Like any other regular array, c can be modified. Everywhere else: it generates an: unnamed array of char What is the type of string literals in C and C++? with static storage that gives UB if modified So when you write: char *c = "abc"; This is similar to: /* __unnamed is magic because modifying it gives UB. */ static char __unnamed[] = "abc"; char *c = __unnamed; Note the implicit cast from char[] to char *, which is always legal. Then if you modify c[0], you also modify __unnamed, which is UB. This is documented at 6.4.5 "String literals": 5 In translation phase 7, a byte or code of value zero is appended to each multibyte character sequence that results from a string literal or literals. The multibyte character sequence is then used to initialize an array of static storage duration and length just sufficient to contain the sequence. For character string literals, the array elements have type char, and are initialized with the individual bytes of the multibyte character sequence [...] 6 It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.

6.7.8/32“初始化”给出了一个直接的例子:

EXAMPLE 8: The declaration char s[] = "abc", t[3] = "abc"; defines "plain" char array objects s and t whose elements are initialized with character string literals. This declaration is identical to char s[] = { 'a', 'b', 'c', '\0' }, t[] = { 'a', 'b', 'c' }; The contents of the arrays are modifiable. On the other hand, the declaration char *p = "abc"; defines p with type "pointer to char" and initializes it to point to an object with type "array of char" with length 4 whose elements are initialized with a character string literal. If an attempt is made to use p to modify the contents of the array, the behavior is undefined.

GCC 4.8 x86-64 ELF实现

计划:

#include <stdio.h>

int main(void) {
    char *s = "abc";
    printf("%s\n", s);
    return 0;
}

编译和反编译:

gcc -ggdb -std=c99 -c main.c
objdump -Sr main.o

输出包含:

 char *s = "abc";
8:  48 c7 45 f8 00 00 00    movq   $0x0,-0x8(%rbp)
f:  00 
        c: R_X86_64_32S .rodata

结论:GCC将char* it存储在.rodata部分，而不是在.text中。

如果我们对char[]做同样的操作:

 char s[] = "abc";

我们获得:

17:   c7 45 f0 61 62 63 00    movl   $0x636261,-0x10(%rbp)

因此它被存储在堆栈中(相对于%rbp)。

但是请注意，默认的链接器脚本将.rodata和.text放在同一个段中，该段有执行权限，但没有写权限。这可以观察到:

readelf -l a.out

它包含:

 Section to Segment mapping:
  Segment Sections...
   02     .text .rodata

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。

这些答案大部分都是正确的，但为了更清楚一点……

人们所说的“只读内存”是ASM术语中的文本段。它是内存中加载指令的同一个地方。出于安全等明显的原因，这是只读的。当创建一个初始化为字符串的char*时，字符串数据被编译到文本段中，程序初始化指向文本段的指针。所以如果你想改变它，就死定了。段错误。

当作为数组编写时，编译器将初始化的字符串数据放在数据段中，这与全局变量等存在的位置相同。这个内存是可变的，因为数据段中没有指令。这一次，当编译器初始化字符数组(仍然只是一个char*)时，它指向的是数据段而不是文本段，您可以在运行时安全地更改文本段。