我是C编程新手。我在使用C中的open()函数写入文件时遇到问题,为清楚起见,这是我的代码
#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该程序时,该程序将按预期运行。
首次运行:
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,它将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,打开文件时出现错误。
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
如何解决文件创建中的权限问题?
答案 0 :(得分:3)
程序首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
您应该像这样使用|
组合模式:
fd = open(datafile, O_CREAT|O_RDWR|O_APPEND, S_IRUSR | S_IWUSR);
您可以使用open
来检查应将哪些参数传递给man open
,在我的系统上,它会显示以下内容(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);
答案 1 :(得分:1)
open()
的参数不正确。 documentation of open()
说:
int open(const char * pathname,int flags,mode_t mode);
O_APPEND
是一个标志,应该与其他两个符号进行或运算以产生flags
参数。
S_IRUSR
和S_IWUSR
是权限。应该对它们进行或运算以产生mode
参数。
总而言之,对open()
的呼叫应为:
fd = open(datafile, O_CREAT | O_RDWR | O_APPEND, S_IRUSR | S_IWUSR);