c++ - 为什么不能输出另外一个文件的字符串？

Question

// file1.c
char str[] = "abcd";

// file2.c
#include <stdio.h>
extern char *str;

int main()
{
    str[0] = 'A';
    printf("%s", str);
}

编译正常，运行崩溃，为什么？

Answer 1

There is a difference between

char str[] and char *str. str[] is the name of the array and is a label. Although str is the first address of the array, str[] itself does not occupy space and is just a symbol. And char *str takes up 4 bytes (8 bytes for 64 bits).

Assuming that on a 32-bit machine, the linker will change the 4 bytes starting from str[] (i.e. str[0], str[1], str[2], str[3]) when linking. The value stored inside is copied to the 4-byte memory space where *str is located. This means that the value of the 4-byte memory space occupied by *str becomes a concatenated ascii value of "abcd". value. Therefore, *str points to an unknown area, and reading and writing to this unknown area may trigger segmentation fault.

If you execute printf("%d",str); on *str, the value should be (little endian)

97 + 98*256 + 99*256*256 + 100*256*256*256

Addition: There is an error in the above explanation. *strIn fact, 4 bytes of storage space is not allocated because it is extern. To put it figuratively, *str is "a hot face but a cold butt" 贴 to str[]. The error mechanism is the same.

Answer 2

extern char str[];
The pointer is not equal to the array name.