4

我收到以下 C++ 代码的“分段错误”:

#include <cstdio>

int main(int, char**) {
  FILE *fp;
  fp = fopen("~/work/dog.txt", "w");
  fprintf(fp, "timer, timer3, timer5, timer6, timer7");
  fclose(fp);
}
4

3 回答 3

9

您的路径无效并且永远不会起作用,因此fopen设置fpNULL并且您会遇到段错误。提示:~字符由 shell 扩展,你不能在fopen.

您尝试做的正确、安全的实现可能如下所示。这是经过测试的。这也是理智的人不使用 C 编写的原因,除非他们没有其他方法:)

// main.cpp
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <unistd.h>

int main(int, char**)
{
    const char * home = getenv("HOME");
    if (home == NULL || strlen(home) == 0 || access(home, F_OK) == -1) abort();
    const char * subPath = "/work/dog.txt";
    char * path = (char*)malloc(strlen(home) + strlen(subPath) + 1);
    if (path == NULL) abort();
    strcpy(path, home);
    strcat(path, subPath);
    FILE *fp = fopen(path, "w");
    if (fp != NULL) {
        fprintf(fp, "timer, timer3, timer5, timer6, timer7");
        fclose(fp);
    }
    free(path);
}
于 2013-06-12T16:46:01.247 回答
1

一些东西:

  • 您需要在使用 fp 之前检查它是否为 NULL,否则只要找不到文件,就会出现段错误。

  • 您需要在将完整路径传递给 fopen 之前解析完整路径(fopen 不知道如何处理“~”)

例子:

FILE *fp = NULL;
char path[MAX];
char *home = getenv ("HOME");
if ( home ) 
{
    snprintf(path, sizeof(path), "%s/work/dog.txt", home);
    // now use path in fopen
    fp = fopen(path, "w");

    if ( fp )
    {
        fprintf(fp, "timer, timer3, timer5, timer6, timer7");
        fclose(fp);
    }
    else
    {
        std::cout << "your dog is missing" << std::endl;
    }
else
{
    std::cout << "You are homeless" << std::endl;
}
于 2013-06-12T17:02:24.773 回答
0

发生段错误是您尝试打开的文件不存在。这与Qt无关。

测试 'fp' 的无效性并正确处理错误。就像是

FILE *fp = fopen("/path/to/work/dog.txt", "w");
if (fp == NULL)
{
    printf("File does not exist.\n");
    // throw exception or whatever.
}
于 2013-06-12T16:37:59.110 回答