1

所以基本上，这就是产生我的代码segmentation fault

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

    #define CORRUPT_KEY_FILE    "s64581-source-key-corrupt.bin" /* ContainsKey and IV */

    #define DEBUG               1

    /**
     * Reads content of a given file and writes it into a buffer
     * 
     * @param filename Path of the file to be read
     * @param buffer Pointer to the buffer to write into
     * @param length Variable to store the filelength into
     * 
     * @return EXIT_FAILURE if an error occurred opening th File, otherwise EXIT_SUCCESS
     */
    int readFiletoBuffer(const char* filename, unsigned char* buffer, int* length) {

        int i;

        if(DEBUG) printf("\n\nFunction readFileToBuffer(..) called\n");

        FILE *file = fopen(filename, "rb");
        if(!file) {
            printf("Error opening file %s", filename);
            return EXIT_FAILURE;
        }

        fseek(file, 0 , SEEK_END);
        *length = ftell(file);
        rewind(file);

        if(DEBUG) printf("\tLength of File %s: %d\n",filename ,*length);

        buffer = (unsigned char*) malloc(*length+1);

        fread(buffer, *length, 1, file);
        fclose(file);

        for(i=0; i<*length; i++) printf("%d:\t%X\n",14, buffer[14]);

        return EXIT_SUCCESS;
    }

    int main(int argc, char **argv)
    {
        unsigned char *tmp_char;
        int tmp_length, i;
        tmp_char=NULL;

        readFiletoBuffer(CORRUPT_KEY_FILE, tmp_char, &tmp_length);

        for(i=0; i<tmp_length; i++) printf("%d:\t%X\n",i, tmp_char[i]);

        return 0;
    }

注意打印缓冲区内容的两个 for 循环。函数内部的那个工作正常，main-function 中的那个产生一个segmentations fault. 为什么会这样，在函数内分配所需内存时如何解决这个问题readFiletoBuffer？

任何帮助表示赞赏！:D

4

2 回答 2

1

问题是，当你readFiletoBuffer用参数调用你的函数时tmp_char，这个参数将被复制！调用此函数后，您的 tmp_char 为 NULL！

您需要使用指向此参数的指针来更改它，如下所示：

tmp_char=NULL;
readFiletoBuffer(CORRUPT_KEY_FILE, &tmp_char, &tmp_length);

所以你的函数看起来像这样：

int readFiletoBuffer(const char* filename, unsigned char** buffer, int* length)
于 2013-06-24T10:18:19.593 回答
1

要从函数中获取所需的 char 数组，您必须将 char** 数组作为参数传递。这样，您的 char* 指针将被返回。

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

#define CORRUPT_KEY_FILE    "s64581-source-key-corrupt.bin" /* ContainsKey and IV */

#define DEBUG               1

/**
 * Reads content of a given file and writes it into a buffer
 * 
 * @param filename Path of the file to be read
 * @param buffer Pointer to the buffer to write into
 * @param length Variable to store the filelength into
 * 
 * @return EXIT_FAILURE if an error occurred opening th File, otherwise EXIT_SUCCESS
 */
int readFiletoBuffer(const char* filename, unsigned char** buffer, int* length) {

    int i;

    if(DEBUG) printf("\n\nFunction readFileToBuffer(..) called\n");

    FILE *file = fopen(filename, "rb");
    if(!file) {
        printf("Error opening file %s", filename);
        return EXIT_FAILURE;
    }

    fseek(file, 0 , SEEK_END);
    *length = ftell(file);
    rewind(file);

    if(DEBUG) printf("\tLength of File %s: %d\n",filename ,*length);

    *buffer = (unsigned char*) malloc(*length+1);

    fread(*buffer, *length, 1, file);
    fclose(file);

    for(i=0; i<*length; i++) printf("%d:\t%X\n",14, buffer[14]);

    return EXIT_SUCCESS;
}

int main(int argc, char **argv)
{
    unsigned char *tmp_char;
    int tmp_length, i;
    tmp_char=NULL;

    readFiletoBuffer(CORRUPT_KEY_FILE, &tmp_char, &tmp_length);

    for(i=0; i<tmp_length; i++) printf("%d:\t%X\n",i, tmp_char[i]);

    return 0;
}

这是我的建议。

于 2013-06-24T10:18:42.450 回答