C++ 使用结构 Stat()

Using Struct Stat()

提问by Dan Snyder

I'm trying to figure out how exactly to use stat() to capture information about a file. What I need is to be able to print several fields of information about a file. So..

我试图弄清楚如何使用 stat() 来捕获有关文件的信息。我需要的是能够打印有关文件的多个信息字段。所以..

 #include <iostream>
 #include <sys/types.h>
 #include <sys/stat.h>
 #include <fcntl.h>
 using namespace std;

 int main() {
     struct stat buf;
     stat("file",&buf);
               ...
     cout << st_dev << endl;
     cout << st_ino << endl;
     cout << st_mode << endl;
     cout << st_nlink << endl;
     cout << st_uid << endl;
     cout << st_gid << endl;
     cout << st_rdev << endl;
     cout << st_size << endl;
     cout << st_blksize << endl;
     cout << st_blocks << endl;
     cout << st_atime << endl;
     cout << st_mtime << endl;
     cout << st_ctime << endl;
     ...
 }

I'm thoroughly confused about how to do this. Why is &buf a parameter to stat? I don't care about storing this information in memory, I just need the outputted fields within my c++ program. How do I access the information contained in the struct? Is buf actually supposed to contain the returned information from stat()?

我对如何做到这一点感到非常困惑。为什么 &buf 是 stat 的参数？我不在乎将这些信息存储在内存中，我只需要我的 c++ 程序中的输出字段。如何访问结构中包含的信息？buf 实际上应该包含从 stat() 返回的信息吗？

回答by Tyler McHenry

Yes, bufis being used here as an out-parameter. The results are stored in bufand the return value of statis an error code indicating if the statoperation succeeded or failed.

是的，buf这里用作输出参数。结果存储在buf，返回值stat是一个错误代码，指示stat操作是成功还是失败。

It is done this way because statis a POSIX function, designed for C, which does not support out-of-band error reporting mechanisms like exceptions. If statreturneda struct, then it would have no way to indicate errors. Using this out-parameter method also allows the caller to choose where they want to store the results, but that's a secondary feature. It's perfectly fine to pass the address of a normal local variable, just like you have done here.

这样做是因为它stat是一个为 C 设计的 POSIX 函数，它不支持像异常这样的带外错误报告机制。如果stat返回一个结构体，那么它将无法指示错误。使用这个 out-parameter 方法还允许调用者选择他们想要存储结果的位置，但这是次要功能。传递普通局部变量的地址完全没问题，就像你在这里所做的那样。

You access the fields of a struct like you would any other object. I presume you are at least familar with object notation? E.g. the st_devfield within the statstruct called bufis accessed by buf.st_dev. So:

您可以像访问任何其他对象一样访问结构体的字段。我想你至少熟悉对象符号？例如，被调用st_dev的stat结构中的字段buf由 访问buf.st_dev。所以：

cout << buf.st_dev << endl;

etc.

等等。

回答by Jon Bringhurst

For another project, I've whipped up a little function that does something similiar to what you need. Take a look at sprintstatf.

对于另一个项目，我创建了一个小函数，可以完成与您需要的类似的功能。看看sprintstatf。

Here's an example of usage:

下面是一个使用示例：

#include <sys/stat.h>
#include <stdlib.h>
#include <stdio.h>

#include "sprintstatf.h"

int
main(int argc, char *argv[])
{
    char *outbuf = (char *)malloc(2048 * sizeof(char));
    struct stat stbuf;
    char *fmt = \
        "st_atime (decimal) = \"%a\"\n"
        "st_atime (string)  = \"%A\"\n"
        "st_ctime (decimal) = \"%c\"\n"
        "st_ctime (string)  = \"%C\"\n"
        "st_gid   (decimal) = \"%g\"\n"
        "st_gid   (string)  = \"%G\"\n"
        "st_ino             = \"%i\"\n"
        "st_mtime (decimal) = \"%m\"\n"
        "st_mtime (string)  = \"%M\"\n"
        "st_nlink           = \"%n\"\n"
        "st_mode  (octal)   = \"%p\"\n"
        "st_mode  (string)  = \"%P\"\n"
        "st_size            = \"%s\"\n"
        "st_uid             = \"%u\"\n"
        "st_uid             = \"%U\"\n";

    lstat(argv[1], &stbuf);

    sprintstatf(outbuf, fmt, &stbuf);
    printf("%s", outbuf);

    free(outbuf);
    exit(EXIT_SUCCESS);
}

/* EOF */

回答by Ram

This question may be way to old to comment but i am posting this as a reference

这个问题可能已经过时了，但我将其发布为参考

To get a good understanding about stat() function ,the reason for passing the stat reference and more importantly error handling are explained good in the below link

为了更好地理解 stat() 函数，传递 stat 引用的原因和更重要的错误处理在下面的链接中有很好的解释

stat - get file status

stat - 获取文件状态

回答by unwind

You have several errors in your code:

您的代码中有几个错误：

• You need &buf, with a single 'f'.
• You need to say e.g. buf.st_devwhen printing, since st_devis a field in the struct variable.

• 你需要&buf, 带有一个'f'。
• 你需要说例如buf.st_dev在打印时，因为st_dev是结构变量中的一个字段。

Since bufis a local variable on the stack, you're not "saving the values to memory" permanently, it's just as long as that variable is in-scope.

由于buf是堆栈上的局部变量，因此您不会永久“将值保存到内存中”，只要该变量在范围内即可。

This is how you return multiple values, typically, in C and C++. You pass a pointer to a structure, and the function being called fills in the structure with the values it has computed for you.

这就是您通常在 C 和 C++ 中返回多个值的方式。你传递一个指向结构的指针，被调用的函数用它为你计算的值填充结构。

回答by RC.

bufis the structure that stat loads with the information about the file you pass in the first parameter. You pass &bufhere b/c you have bufallocated on the stack as a local variable and you must pass a pointer to the stat function to enable it to load the data.

buf是 stat 加载的结构，其中包含有关您在第一个参数中传递的文件的信息。您&buf在此处传递已buf在堆栈上分配的b/c作为局部变量，并且您必须将指针传递给 stat 函数以使其能够加载数据。

All variables of st_*are part of the struct stat object and thus must be accessed via your local bufvariable as buf.st_uid, etc.

的所有变量st_*都是 struct stat 对象的一部分，因此必须通过本地buf变量 asbuf.st_uid等访问。