查找目录的大小 [英] Finding the size of a directory

问题描述

以下是这样一个函数的伪代码：跟随一个递归方法请告诉我是否还有其他方法。

  int directorySize（DirectoryHandle dh）
 {
 int size = 0; 
 if（！dh）
 {
 DirectoryHandle dh1 = directoryOpen（Directory_path）; 
} 
 else 
 {
 dh1 = dh; 
} 
 
 while（dh1）
 {
 if（TRUE = IsDirectory（dh1））
 {
 size + = directorysize（dh1 ）; 
} 
 else if（TRUE == IsFile（dh1））
 {
 FileHandle fh = dh1; 
 while（EOF！= fh）
 {
 size ++; 
} 
} 
} 
} 
  

解决方案

使用nftw的典型例子：

请注意，随着面试问题的出现，他们可能会希望看到您考虑

• 遍历顺序


• 权限（无法访问的子文件夹等）


• ondisk vs.表面大小


• 符号链接，硬链接（树外）重复计数？


• 稀疏文件


• 表现

以下代码以务实的方式解决了大部分问题：

。

  #define _XOPEN_SOURCE 500 
 #include< ftw.h> 
 #include< stdio.h> 
 #include< stdlib.h> 
 #include< stdint.h> 
 
 static uintmax_t total = 0ul; 
 static uintmax_t files = 0ul; 
 static uintmax_t directories = 0ul; 
 static uintmax_t symlinks = 0ul; 
 static uintmax_t uncessible = 0ul; 
 static uintmax_t blocks512 = 0ul; 
 
 static int 
 display_info（const char * fpath，const struct stat * sb，
 int tflag，struct FTW * ftwbuf）
 {
 switch tflag）
 {
 case FTW_D：
 case FTW_DP：directories ++;打破; 
 case FTW_NS：
 case FTW_SL：
 case FTW_SLN：symlinks ++;打破; 
 case FTW_DNR：不可访问++;打破; 
 case FTW_F：files ++;打破; 
} 
 total + = sb-> st_size; 
 blocks512 + = sb-> st_blocks; 
 return 0; / *告诉nftw（）继续* / 
} 
 
 int 
 main（int argc，char * argv []）
 {
 int flags = FTW_DEPTH | FTW_MOUNT | FTW_PHYS; 
 
 if（nftw（（argc <2）？。：argv [1]，display_info，20，flags）== -1）
 {
 perror nftw）; 
 exit（EXIT_FAILURE）; 
} 
 
 printf（总大小：％7jd\\\
，总计）; 
 printf（在％jd文件和％jd目录（％jd符号链接和％jd不可访问的目录）\\\
，文件，目录，符号链接，不可访问）; 
 printf（磁盘上的大小％jd * 512b =％jd\\\
，blocks512，blocks512< 9）; 
 
 exit（EXIT_SUCCESS）; 
} 
  

这是张贴为以前获得目录大小和大小的最快方法。典型输出：

 总大小：28433001733 
在878794个文件和87047个目录（73318个符号链接和0个不可访问的目录）
磁盘上的大小59942192 * 512b = 30690402304 
  

I got this question in a Cisco interview: write a function to find the size of a directory?

Following is the pseudocode for such a function, that follows a recursive approach. Please tell me if there can be any other approach also.

int directorySize(DirectoryHandle dh)
{
    int size=0;
    if (!dh)
    {
        DirectoryHandle dh1 = directoryOpen("Directory_path");
    }
    else
    {
        dh1 = dh;
    }

    while (dh1)
    {
        if (TRUE=IsDirectory(dh1))
        {
            size += directorysize(dh1);
        }
        else if (TRUE == IsFile(dh1))
        {
            FileHandle fh = dh1;
            while (EOF != fh)
            {
                size++;
            }
        }
    }
}

解决方案

Canonical example of using nftw:

Note that as interview questions go, they will probably want to see you thinking about

• traversal order
• permission (inaccessible subfolder etc.)
• size ondisk vs. apparent size
• symlinks, hardlinks (outside the tree? duplicate counting?)
• sparse files
• performance

The following code does address most of these issues in a pragmatic fashion:

.

#define _XOPEN_SOURCE 500
#include <ftw.h>
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

static uintmax_t total        = 0ul;
static uintmax_t files        = 0ul;
static uintmax_t directories  = 0ul;
static uintmax_t symlinks     = 0ul;
static uintmax_t inaccessible = 0ul;
static uintmax_t blocks512    = 0ul;

static int
display_info(const char *fpath, const struct stat *sb,
             int tflag, struct FTW *ftwbuf)
{
    switch(tflag)
    {
        case FTW_D:
        case FTW_DP:  directories++;  break;
        case FTW_NS:
        case FTW_SL:
        case FTW_SLN: symlinks++;     break;
        case FTW_DNR: inaccessible++; break;
        case FTW_F:   files++;        break;
    }
    total += sb->st_size;
    blocks512 += sb->st_blocks;
    return 0; /* To tell nftw() to continue */
}

int
main(int argc, char *argv[])
{
    int flags = FTW_DEPTH | FTW_MOUNT | FTW_PHYS;

    if (nftw((argc < 2) ? "." : argv[1], display_info, 20, flags) == -1)
    {
        perror("nftw");
        exit(EXIT_FAILURE);
    }

    printf("Total size: %7jd\n", total);
    printf("In %jd files and %jd directories (%jd symlinks and %jd inaccessible directories)\n", files, directories, symlinks, inaccessible);
    printf("Size on disk %jd * 512b = %jd\n", blocks512, blocks512<<9);

    exit(EXIT_SUCCESS);
}

This was posted as Fastest ways to get a directory Size and Size on disk before. Typical output:

Total size: 28433001733
In 878794 files and 87047 directories (73318 symlinks and 0 inaccessible directories)
Size on disk 59942192 * 512b = 30690402304

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