opendir error?

Question

I get an error with this istruction:

  dp = opendir ("%APPDATA%/.");

  output: 
  Couldn't open directory: Mo such file or directory.

but I don't get an erro with this istruction:

dp = opendir ("C:/Users/xrobot/AppData/.");

output:
.
..
Local
LocalLow
Roaming

Why ?

Daniel Fischer · Accepted Answer · 2012-04-20T12:43:55.603

7

opendir doesn't expand meta variables like %APPDATA%, the shell does. So such things work from the command line, but not from a program. In your program, you have to use an absolute or relative path.

You can probably obtain the required path with getenv(),

const char *appData = getenv("APPDATA");
if (appData) {
    dp = opendir(appData);
} else {
    /* die or recover */
}

edited Apr 20 '12 at 12:43

answered Apr 20 '12 at 12:29

Daniel Fischer

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I'm no Windows expert, but you'd probably have to use `getenv()` for that. Try `const char* appData = getenv("APPDATA"); dp = opendir(appData);` – Component 10 Apr 20 '12 at 12:40
I don't know about C++, in C you could use `getenv` to look up the value of `APPDATA`. Well, at least on *nixish systems, not sure about Windows. – Daniel Fischer Apr 20 '12 at 12:40

score 2 · Answer 2 · answered Apr 20 '12 at 12:29

2

Because the first opendir is LITERALLY trying to open the directory %APPDATA%/..

answered Apr 20 '12 at 12:29

Ed Heal

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opendir error?

2 Answers2