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?
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()的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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