linux下tree、命令的用法及实现代码 |sunstars2009918
Linux下有这样一个命令,可以把当前目录下的所有文件和子文件以tree的方式显示出来,看下效果[*][*][www.linuxidc.com@localhost test]$ tree[*].[*]|-- A[*]|-- B[*]|-- C[*]`-- test2[*] |-- D[*] |-- E[*] `-- F[*]
[*]3 directories, 4 files[*]$[*]
自己用递归方式用C实现了下,效果如下: [*]
[*][www.linuxidc.com@localhost test]$ ./a.out [*]./test[*]A[*]a.out[*]B[*]C[*]+test2[*] F[*] +D[*] +E[*]$[*]
这里+号表示directory.
下面是源码:
#include <iostream>
#include <sys/stat.h>
#include <dirent.h>
#include <vector>
using namespace std;
int showConsoleDir(char* path, int cntFloor) {
DIR* dir;
DIR* dir_child;
struct dirent* dir_ent;
if ((dir = opendir(path))==NULL) { //open current directory
cout<<"open dir failed!"<<endl;
return -1;
}
while ((dir_ent = readdir(dir))!=NULL) {
if ((dir_ent->d_name == '.') || (strcmp(dir_ent->d_name, "..") ==0)){ //if . or .. directory continue
continue;
}
char tName;
memset(tName, 0, 10000);
snprintf(tName,sizeof(tName),"%s/%s",path,dir_ent->d_name);
if ((dir_child = opendir(tName))!=NULL){//if have a directory
int t = cntFloor;
while (t--) {
cout<<"";
}
cout<<"+"<<dir_ent->d_name<<endl;
showConsoleDir(tName, cntFloor+1);
}
else
{
int t = cntFloor;
while (t--) {
cout<<"";
}
cout<<dir_ent->d_name<<endl;
}
}
}
int main(int argc, char* argv[]){
int cntFloor=1;
showConsoleDir("./", cntFloor);
}
作者：sunstars2009918 发表于2011-12-16 8:05:16 原文链接
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