为什么初始化结构后strcpy中存在分段错误？

Question

我似乎无法弄清楚为什么strcpy在这段代码中会出现分段错误。 它应该很简单：

    typedef struct message {
        char *buffer;
        int length;
    } message_t;

    int main () {
        char* buf = "The use of COBOL cripples the mind; its "
                    "teaching should, therefore, be regarded as a criminal "
                    "offense. -- Edsgar Dijkstra";
        message_t messageA = {"", 130};
        message_t *message = &messageA;
        strcpy(message->buffer, "buf");
        printf("Hello\n");
    }

编辑： strcpy(message->buffer, "buf")应该是strcpy(message->buffer, buf)没有 "" 引号

编辑：感谢您的评论：这已通过 malloc'ing message->buffer 为 buf 腾出空间来解决：

    message_t messageA = {"", 130};
    message_t *message = &messageA;
    message->buffer = malloc(122);
    strcpy(message->buffer, buf);
    printf("Hello\n");

Answer 1

这里有几点需要注意。

当您声明存储数据的指针时，您可以直接在声明中分配（通常用于小字符串而不是大字符串），或者您应该使用动态 memory 分配函数分配 memory

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct message {
    char *buffer;
    int length;
} message_t;

int main () {
    char* buf = "The use of COBOL cripples the mind; its "
                "teaching should, therefore, be regarded as a criminal "
                "offense. -- Edsgar Dijkstra";
    message_t messageA = {"", 130};
    message_t *message = &messageA;

    //allocate memory to buffer length of buf, + 1 for \0 character
    message->buffer = malloc( strlen(buf) + 1 );
    // check allocated memory is success or not
    if ( message->buffer )
    {
        strcpy(message->buffer, buf);
        printf("buffer = %s\n",message->buffer );
        //free the malloc'ed memory to avoid memory leaks
        free(message->buffer);
        //make the pointer NULL, to avoid dangling pointer
        message->buffer = NULL;
    }
    else
    {
        printf("malloc failed\n");
    }
    printf("Hello\n");
    return 0;
}

Answer 2

message_t messageA = {"", 130};

在这里，您初始化messageA.buffer = "" 。 它是一个字符串文字。 因此，您不能修改存储在其中的字符串。 如果你试图修改它，你会得到分段错误。

message_t *message = &messageA;
strcpy(message->buffer, buf);

在这里，您正在修改字符串文字message->buffer 。 这就是您遇到分段错误的原因。
请访问此问题Modifying a string literal

Answer 3

尝试使用message->buffer = strdup(buf) ; 为您执行malloc和strlen计算。