c - C 中 open() 函数中的 "Permission denied"

Question

我是 C 编程的新手。我在使用 C 中的 open() 函数写入文件时遇到问题，为了清楚起见，这里是我的代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <fcntl.h>
#include <sys/stat.h>
#include <unistd.h>

void usage(char *prog_name, char *filename){
    printf("Usage: %s <data to add to %s> \n",prog_name, filename);
    exit(0);
}

void fatal(char *);                          
void *errchck_malloc(unsigned int);     

int main(int argc, char *argv[]){
    int fd; // File descriptor
    char *buffer, *datafile;

    buffer = (char *) errchck_malloc(100);
    datafile = (char *) errchck_malloc(20);
    strcpy(datafile, "./simplenote.txt");

    if (argc < 2)
        usage(argv[0], datafile);

    strcpy(buffer, argv[1]);

    printf("[DEBUG] buffer @ %p: \'%s\'\n", buffer, buffer);
    printf("[DEBUG] datafile @ %p: \'%s\'\n", datafile, datafile);

    strncat(buffer, "\n", 1);

    // Open file 
    fd = open(datafile, O_CREAT|O_RDWR,O_APPEND, S_IRUSR, S_IWUSR);
    if(fd == -1)
        fatal("in main() while opening file");
    printf("[DEBUG] file descriptor is %d\n", fd);

    // Writing data to file
    if(write(fd, buffer, strlen(buffer))==-1)
        fatal("in main() while writing buffer to file");

    // Closing file
    if(close(fd) == -1)
        fatal("in main() while closing file");

    printf("Note saved\n");
    free(buffer);
    free(datafile);
}

// fatal(): Function to display error message then exit
void fatal(char *message){
    char err_msg[100];

    strcpy(err_msg, "[!!] Fatal Error ");
    strncat(err_msg, message, 83);
    perror(err_msg);
    exit(-1);
}

// errchck_malloc(): An error check malloc wrapper function
void *errchck_malloc(unsigned int size){
    void *ptr;
    ptr = malloc(size);
    if(ptr == NULL)
        fatal("in errchck_malloc() on memory allocation");
    return ptr;
}

当我第一次尝试执行程序时，程序按预期运行。

第一次运行:

user: ./simplenote "Hello, again"
[DEBUG] buffer @ 0x7fafcb4017a0: 'Hello again'
[DEBUG] datafile @ 0x7fafcb401810: './simplenote.txt'
[DEBUG] file descriptor is 3
Note saved

当我尝试打开文件并查看文本时，出现权限被拒绝错误。当我尝试使用 sudo 打开文件时，它打开并且文本在文件中。当我第二次运行该程序时，由于权限问题，我在打开文件时遇到错误。

第二次运行:

user: ./simplenote "just checking if it is still working"                                                           
[DEBUG] buffer @ 0x7face4c017a0: 'just checking if it is still working'
[DEBUG] datafile @ 0x7face4c01810: './simplenote.txt'
[!!] Fatal Error in main() while opening file: Permission denied

如何解决文件创建的权限问题？

Answer 1

程序第一次运行时创建的文件具有不允许您追加的权限

stat simplenote.txt 
  File: simplenote.txt
  Size: 5           Blocks: 8          IO Block: 4096   regular file
Device: 2fh/47d Inode: 32810078    Links: 1
Access: (2000/------S---)  Uid: ( 1000/ user)   Gid: ( 1000/ user)
Access: 2020-01-05 19:29:34.317872734 +0100
Modify: 2020-01-05 19:29:34.317872734 +0100
Change: 2020-01-05 19:29:34.317872734 +0100

你应该像这样使用 | 组合模式:

fd = open(datafile, O_CREAT|O_RDWR|O_APPEND, S_IRUSR | S_IWUSR);

你可以使用 man open 检查你应该传递给 open 的参数，在我的系统上它显示了这样的东西(修剪到重要部分):

   int open(const char *pathname, int flags);
   int open(const char *pathname, int flags, mode_t mode);

   int creat(const char *pathname, mode_t mode);

   int openat(int dirfd, const char *pathname, int flags);
   int openat(int dirfd, const char *pathname, int flags, mode_t mode);