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

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。

其他回答

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。

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.

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。

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));

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