為什么初始化結構后strcpy中存在分段錯誤？

Question

我似乎無法弄清楚為什么strcpy在這段代碼中會出現分段錯誤。 它應該很簡單：

    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");
    }

編輯： strcpy(message->buffer, "buf")應該是strcpy(message->buffer, buf)沒有 "" 引號

編輯：感謝您的評論：這已通過 malloc'ing message->buffer 為 buf 騰出空間來解決：

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

Answer 1

這里有幾點需要注意。

當您聲明存儲數據的指針時，您可以直接在聲明中分配（通常用於小字符串而不是大字符串），或者您應該使用動態 memory 分配函數分配 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};

在這里，您初始化messageA.buffer = "" 。 它是一個字符串文字。 因此，您不能修改存儲在其中的字符串。 如果你試圖修改它，你會得到分段錯誤。

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

在這里，您正在修改字符串文字message->buffer 。 這就是您遇到分段錯誤的原因。
請訪問此問題Modifying a string literal

Answer 3

嘗試使用message->buffer = strdup(buf) ; 為您執行malloc和strlen計算。