C++ c++;,linux,如何解释'stat64'中文件的模式?
在那里,我试图确定一个文件是文件夹还是文件夹中的文件C++ c++;,linux,如何解释'stat64'中文件的模式?,c++,linux,C++,Linux,在那里,我试图确定一个文件是文件夹还是文件夹中的文件 struct dirent **name_list; int n, i; n = scandir(".", &name_list, NULL, alphasort); for(i=0;i<n;i++){ struct stat64 stat_list stat64(name_list[i]->d_name, &stat_list); cout << stat_list.st_mod
struct dirent **name_list;
int n, i;
n = scandir(".", &name_list, NULL, alphasort);
for(i=0;i<n;i++){
struct stat64 stat_list
stat64(name_list[i]->d_name, &stat_list);
cout << stat_list.st_mode << endl;
}
struct dirent**name\u列表;
int n,i;
n=scandir(“.”,&name\u list,NULL,alphasort);
对于(i=0;id\u名称和统计列表);
我会说:)
我会说:)
stat
的man
页面有一个表,其中列出了每个标志的含义
S_IFMT 0170000 bit mask for the file type bit fields
S_IFSOCK 0140000 socket
S_IFLNK 0120000 symbolic link
S_IFREG 0100000 regular file
S_IFBLK 0060000 block device
S_IFDIR 0040000 directory
S_IFCHR 0020000 character device
S_IFIFO 0010000 FIFO
S_ISUID 0004000 set UID bit
S_ISGID 0002000 set-group-ID bit (see below)
S_ISVTX 0001000 sticky bit (see below)
S_IRWXU 00700 mask for file owner permissions
S_IRUSR 00400 owner has read permission
S_IWUSR 00200 owner has write permission
S_IXUSR 00100 owner has execute permission
S_IRWXG 00070 mask for group permissions
S_IRGRP 00040 group has read permission
S_IWGRP 00020 group has write permission
S_IXGRP 00010 group has execute permission
S_IRWXO 00007 mask for permissions for others (not in group)
S_IROTH 00004 others have read permission
S_IWOTH 00002 others have write permission
S_IXOTH 00001 others have execute permission
问题中模式的数字表示形式输出为十进制,但是如果将其转换为八进制,则位字段更有意义
文件的33188
转换为0o100644
,这意味着它是一个常规文件,所有者只读/写,组/其他只读<目录的code>16877
转换为0o40755
,这意味着它是一个具有所有所有者权限的目录,并且对组和其他对象具有读取/执行权限。用于stat
的man
页面有一个每个标志的含义表
S_IFMT 0170000 bit mask for the file type bit fields
S_IFSOCK 0140000 socket
S_IFLNK 0120000 symbolic link
S_IFREG 0100000 regular file
S_IFBLK 0060000 block device
S_IFDIR 0040000 directory
S_IFCHR 0020000 character device
S_IFIFO 0010000 FIFO
S_ISUID 0004000 set UID bit
S_ISGID 0002000 set-group-ID bit (see below)
S_ISVTX 0001000 sticky bit (see below)
S_IRWXU 00700 mask for file owner permissions
S_IRUSR 00400 owner has read permission
S_IWUSR 00200 owner has write permission
S_IXUSR 00100 owner has execute permission
S_IRWXG 00070 mask for group permissions
S_IRGRP 00040 group has read permission
S_IWGRP 00020 group has write permission
S_IXGRP 00010 group has execute permission
S_IRWXO 00007 mask for permissions for others (not in group)
S_IROTH 00004 others have read permission
S_IWOTH 00002 others have write permission
S_IXOTH 00001 others have execute permission
问题中模式的数字表示形式输出为十进制,但是如果将其转换为八进制,则位字段更有意义
文件的
33188
转换为0o100644
,这意味着它是一个常规文件,所有者只读/写,组/其他只读<代码>16877用于将目录转换为0o40755
,这意味着它是一个具有所有所有者权限的目录,并且对组和其他文件具有读取/执行权限。是的,很抱歉,我不认为使用“man”来表示这一点。。。几乎每次当我使用man+函数时,我都一无所获。。。但好吧,我同意在这种情况下,告诉我RTFM是公平的。如果你在man
方面运气不好,我建议你使用一个好的搜索引擎,比如说,是的,我很抱歉我不认为用“man”来做这个。。。几乎每次当我使用man+函数时,我都一无所获。。。但是,好吧,我同意在这种情况下,告诉我RTFM是公平的。如果你在man
方面运气不好,我建议你使用一个好的搜索引擎,比如,我更新了我的答案,并补充了一些关于标志如何工作的解释。我更新了我的答案,补充了一些关于标志如何工作的解释。