SYSTEM调用执行后的分段错误

som*_*one 0 c segmentation-fault

请告诉我我在做什么是对还是错.如果它是正确的那么为什么我得到分段错误?

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

char* EncodeToBase64(char*);

int main()
{
    char str[24] = {0};
    char *output = NULL;
    output = malloc(56*sizeof(char));
    scanf("%[^\n]s",str);
    output = EncodeToBase64(str);
    printf("Back %s",*output);
    return 0;
}

char* EncodeToBase64(char *str)
{
    char buff[24];
    char *output = NULL;
    output = malloc(56*sizeof(char));
    memset(buff,'0',sizeof(buff));
    printf("EncodeToBase64 Function\n");
    sprintf(buff,"echo %s | openssl enc -base64",str);
    printf("%s\n",buff);
    output = system(buff);
    printf("Back %s",*output);

    return output;
}
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在纠正我的代码后,我又遇到了一个问题,每次我都会在我的字符串中添加这三个字符ô/+...如何获得确切的字符串?

Mar*_*n R 5

system()返回a int,这是进程的退出状态,而不是标准输出.所以分配的返回值system()char *印刷,它没有任何意义.

您可以使用它popen()来启动进程并读取其输出.

示例(使用固定大小的缓冲区,为简洁起见,不进行错误检查):

char * EncodeToBase64(char *str)
{
    char buff[1024], output[1024];

    snprintf(buff, sizeof(buff), "echo %s | openssl enc -base64", str);
    FILE *fp = popen(buff, "r");
    size_t amount = fread(output, 1, sizeof(output) - 1, fp);
    output[amount] = 0; // zero terminate string
    fclose(fp);

    return strdup(output);
}
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用法:

char *b64 = EncodeToBase64("Hello world");
printf("%s", b64);
free(b64);
// Output: SGVsbG8gd29ybGQK
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