linux下tree、命令的用法及實現代碼 .
linux下tree、命令的用法及實現代碼 .
Linux下有這樣一個命令,可以把當前目錄下的所有文件和子文件以tree的方式顯示出來,看下效果01.$ tree
02..
03.|-- A
04.|-- B
05.|-- C
06.`-- test2
07. |-- D
08. |-- E
09. `-- F
10.
11.3 directories, 4 files
12.$ 自己用遞歸方式用C實現了下,效果如下:01.$ ./a.out
02../test
03. A
04. a.out
05. B
06. C
07. +test2
08. F
09. +D
10. +E
11.$ 這裡+號表示directory.
下面是源碼:01.#include
02.#include
03.#include
04.#include
05.using namespace std;
06.
07.int showConsoleDir(char* path, int cntFloor) {
08. DIR* dir;
09. DIR* dir_child;
10. struct dirent* dir_ent;
11.
12. if ((dir = opendir(path))==NULL) { //open current directory
13. cout<<"open dir failed!"<<endl;
14. return -1;
15. }
16. while ((dir_ent = readdir(dir))!=NULL) {
17. if ((dir_ent->d_name == '.') || (strcmp(dir_ent->d_name, "..") ==0)){ //if . or .. directory continue
18. continue;
19. }
20. char tName;
21. memset(tName, 0, 10000);
22. snprintf(tName,sizeof(tName),"%s/%s",path,dir_ent->d_name);
23. if ((dir_child = opendir(tName))!=NULL){ //if have a directory
24. int t = cntFloor;
25. while (t--) {
26. cout<<" ";
27. }
28. cout<<"+"<d_name<<endl;
29. showConsoleDir(tName, cntFloor+1);
30. }
31. else
32. {
33. int t = cntFloor;
34. while (t--) {
35. cout<<" ";
36. }
37. cout<d_name<<endl;
38. }
39. }
40.}
41.
42.int main(int argc, char* argv[]){
43. int cntFloor=1;
44. showConsoleDir("./", cntFloor);
45.
46.
47.} 《解決方案》
學習嘍
