Passing Arguments to execlp()
Posted on
16th Feb 2014 07:03 pm 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);
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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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