reading multiple file names & displaying them

hi,
wondering if someone could explain why the following doesn't work as intended ?
The 2nd for loop should display the file names, but instead it displays the same one, the last one read in.
Now I've tested reading in strings into char array and displaying them again in another for loop, and it works as intended, and I've looked/compared the addresses too but this, has me stumped.

Something I'm missing?

Code:

---

#include <dirent.h>
#include <iostream>
using namespace std;

int main()
{
char *afilelist[100] = { "0" };
DIR                                          *d;
struct dirent *dir[100];
d = opendir(".");

if (d)
{
        for( int x=0; (dir[x] = readdir(d)) != NULL;x++)
        {
                afilelist[x] = dir[x]->d_name;
                cout<<afilelist[x]<<endl; //displays file names properly.
        }
closedir(d);
}

cout<<endl;
for( int x=0;x<10 ; x++)
{
        cout<<afilelist[x]<<endl; //doesn't display file names as per the array ?
}

cin.get();
}

---

Perhaps the system returned pointer for dir[x]->d_name is the same on each call? Print out name and address in your first FOR loop as it runs to see.

If I run the code & return the addresses (and what it returns, is next to it);

First for loop;
0x22fd70 .
0x22fd74 ..
0x22fd78 file1name.txt
0x22fd7c file2name.txt
0x22fd80 fileXname.ext
0x22fd84 fileLast.ext

2nd for loop, the addresses are the same as first, however what it returns is not
0x22fd70 fileLast.ext
0x22fd74 fileLast.ext
0x22fd78 fileLast.ext
0x22fd7c fileLast.ext
It's only displaying/showing the last read in file name.

Now if I do this with some simple char/string array and do the same thing, I don't have the same problem. So I don't understand why the problem is happening with this code.

Looks like you need to copy the directory names into an array of strings, instead of keeping a pointer. I recommend using an array of std::string instead of char *.

Also, while 100 is surely a high enough for any reasonable use of the program, you should still make the program well defined if the user input is not what you expect, namely that there are more than 100 files in the directory. Better predictable, though limited use behavior, than an uncontrolled crash. In this case that means adding a check for x<100 for the first for loop.

By the way: this does not do what you think:

Code:

---

char *afilelist[100] = { "0" }

---

It makes a single pointer that points to the string "0", and the rest are null pointers.

Quote:

---

Originally Posted by King Mir

....while 100 is surely a high enough....

---

My code is just for my own testing/learning.

Quote:

---

Looks like you need to copy the directory names into an array of strings, instead of keeping a pointer. I recommend using an array of std::string instead of char *.

---

Thanks, changing that worked. However, could you/someone possibly explain why the above code doesn't work as intended but the following does;

Code:

---

#include <iostream>
using namespace std;

int main()
{
        char *aChar[15] = { "0" };
        char *cString[] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "end" };

for(int i=0; cString[i]<"end" ; i++)
{
        aChar[i] = cString[i];
}

for(int i=0; cString[i]<"end"; i++)
{
        cout<<aChar[i]<<endl; //displays array as intended
}

cin.get();
}

---

Use strcmp in a while loop like strcmp("end", cString[i]) != 0

If you look at the addresses returned from the readdir() function you posted above, they are only 4 bytes apart. Apparently, readdir() on your system does not guarantee to preserve the contents of the structure across subsequent calls.

And, tell me you realize that this:

Code:

---

for(int i=0; cString[i]<"end" ; i++)

---

is comparing the addresses and not the values.

This works on my system (a Mac). See how it runs on yours.

Code:

---

#include <dirent.h>
#include <iostream>
using namespace std;

int main()
{
int x, y ;
struct dirent * afilelist[100] ; // "0" };
DIR *d;
struct dirent * dir[100];
d = opendir("..");

if (d)
{
        for( x=0; (dir[x] = readdir(d)) != NULL; x++)
        {
                // afilelist[x] = dir[x]->d_name;
                afilelist[x] = dir[x] ;

                cout << afilelist[x] -> d_name << endl; //displays file names properly.
                if (x > 0) cout << "\tPrevious name was " << afilelist[x-1] -> d_name << endl ;
                if (x > 1) cout << "\tPrevious previous name was " << afilelist[x-2] -> d_name << endl ;
        }
closedir(d);
}

cout<<endl;
for( y=0; y < x ; y++)
{
        cout << afilelist[y] -> d_name << endl; //doesn't display file names as per the array ?
}

cin.get();
}

---

Quote:

---

Originally Posted by Dino

If you look at the addresses returned from the readdir() function you posted above, they are only 4 bytes apart. Apparently, readdir() on your system does not guarantee to preserve the contents of the structure across subsequent calls.

This works on my system (a Mac). See how it runs on yours.

---

Works on mine, but again the 2nd for loop only displays the last or most recent filename read in.

Quote:

---

Code:

---

for(int i=0; cString[i]<"end" ; i++)

---

is comparing the addresses and not the values.

---

Really? But if one displays what cString[i] is, it shows/returns the value yes? How is that comparing addresses if "end" is a string literal?

Thanks.

To compare strings, you use strcmp().

Quote:

---

Originally Posted by Tropod

Really? But if one displays what cString[i] is, it shows/returns the value yes? How is that comparing addresses if "end" is a string literal?

Thanks.

---

It is returning the value -- of the char pointer.

A char pointer can be treated like a string; most functions that take or return a char pointer assume that it points to a null terminated set of characters used for input and output. But at the end of the day it's still a pointer to char. If you compare pointers, you are comparing memory locations.

Arrays, including string literals, will decay to pointers in the context of any operator except & (address of).