Passing Arguments to execlp()

Posted on 16th Feb 2014 by admin

I'm writing a program that mimics a unix shell. It's supposed to take commands with arguments and execute them. I'm having trouble passing in the arguments into the execlp call to correctly execute the program. Here's the code I have:

Code: // Construct the command argument array char* args[argsEndIndex]; // Array to hold command arguments int i = 0; // Loop counter to loop through command int j = 0; // Loop counter for argument array while (i <= argsEndIndex) { if (i != 0) { args[j] = cmdParts[i]; j++; } i++; } execlp("/bin/ls", "ls", args, NULL);

cmdParts is a char* array that contains the formatted command inputted. I parsed through the command above this code to insert characters so the pointers point to null terminated strings. That's also how I got argsEndIndex. For example I input this command:

Code: ls -a -l

argsEndIndex gets correctly set to 2 because that's the index of the last argument in the command. 0 is obviously the command itself. Now when I make the execlp call I get:

Code: ls: cannot access sH*?vH*?y: No such file or directory

Which I'm assuming it's accessing garbage data somewhere through the args array but I can't figure out why it's doing it because I'm correctly setting each element of the array.

I can post the whole code if you need it but it's long so I tried to keep this short. Any help is appreciated. Thanks!

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