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

假设字符串是，

char a[] = "string literal copied to stack";
char *p  = "string literal referenced by p";

在第一种情况下，当'a'进入作用域时，文字将被复制。这里'a'是定义在stack上的数组。这意味着字符串将在堆栈上创建，其数据从代码(文本)内存中复制，通常是只读的(这是特定于实现的，编译器也可以将这种只读的程序数据放在可读写内存中)。

在第二种情况下，p是定义在堆栈(本地作用域)上的指针，并引用存储在其他位置的字符串字面量(程序数据或文本)。通常，修改这样的内存不是好的实践，也不鼓励。

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。

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).

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。

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。