How to input data into char array?

Hi,

I'm trying to create a simple unix shell. Right now I'm having problems inputting the command and arguments that a user types in. I need to pass them off to execvp to execute the command. This is what the call to execvp looks like:
int execvp(const char *file, char *const argv[]);

So, I need to get the command as char* and the arguments as char*[]. This is what I tried to do, but got a segmentation fault:
char *c;
cout << "command: ";
cin >> c;
cout << "args: ";
char* args[128];
cin >> args[0] >> args[1];

Any idea how to do this properly? Thanks!

--jumpy

Go back and review your notes on pointers. You are trying to get input into pointers that don't point to anything.

gg

Codeplug,

I looked at the pointer faq on this site, but that didn't help. Can you give me some other resource?

Are you saying that I need to reserve space first using the new operator?

Thanks,

Jumpy

You can reserve space on the stack as well:

Code:

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char input[255];
...
cin >> input;

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gg

>Are you saying that I need to reserve space first using the new operator?
Yes, a pointer has to point to something that you owm before you can safely use it. That means using dynamic memory with new, or by assigning the address of a static of local variable:

Code:

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char *p = new char[SIZE];

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Code:

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char a[SIZE];
char *p = a;

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