simple login program problem

My sample code:

Code:

#include <stdio.h>
#include <string.h>
#define name1 "denny"
#define pass1 "123"

int main(int argc, char *argv[])
{   char name[5];
    char pass[3];

    if(argc != 2)
    {        printf("Usage: login Your_Username Your_Password\n");
             exit(1);
    }
    scanf(argv[1],"%s", &name); // to read username argument
    scanf(argv[2],"%s", &pass; // to read password argument

    if(strcmp(name,name1) == 0 && strcmp(pass,pass1) == 0)
       printf("Correct username and password");

    else
       printf("Wrong username and password");
    return 0;
}

Hi I'm a new user of c programming.
My purpose of doing this programming is letting the user to key in their username and password.

The program must be run from console.

It looks like

D:\c>login
Usage:login Your_Username Your_Password

But the problem is, i do not know how to read the username and the password key in by the user. anyone can fix my program?
I'll be very thankful if someone can help me.

Code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define name1 "denny"
#define pass1 "123"

int main(int argc, char *argv[])
{  
    char * name;
    char * pass;

    if(argc != 3)
    {        
        printf("Usage: login Your_Username Your_Password\n");
         exit(1);
    }

    name = argv[1];
    pass = argv[2];    

    if(strcmp(name,name1) == 0 && strcmp(pass,pass1) == 0)
       printf("Correct username and password\n");
    else
       printf("Wrong username and password\n");

    return 0;
}

thanx a lot for the help.

Login is what I would do with GUI. But it's a matter of taste.

You might want to consider hiding the password when the user types it in. A search of these forums would yield several possible solutions, depending on your environment O/S etc.

Originally Posted by maxorator

Login is what I would do with GUI. But it's a matter of taste.

Right because they've never, ever used text based logins. Ever.


Quzah.

[QUOTE=suckss]My sample code:

Code:

    scanf(argv[1],"%s", &name); // to read username argument
    scanf(argv[2],"%s", &pass; // to read password argument

    }

QUOTE]
After getting the exact answer what u want, Did U try to know what does argc and argv means?

Originally Posted by SKeane

You might want to consider hiding the password when the user types it in. A search of these forums would yield several possible solutions, depending on your environment O/S etc.

nah, i'm quite satisfied with it now. Coz i'm tryin to do bufferoverrun.
I might have a question again later regarding buffer overrun.
anyway thanx for you help SKeane

[QUOTE=maven]

Originally Posted by suckss

My sample code:

Code:

    scanf(argv[1],"%s", &name); // to read username argument
    scanf(argv[2],"%s", &pass; // to read password argument

    }

QUOTE]
After getting the exact answer what u want, Did U try to know what does argc and argv means?

yeah what does argc and agrv means...coz im new to C. i prefer c++ rather than C. i never study c before and it's quite confusing for me to see the codes.

Can i store the username entered by the user and the password to an array instead of storing it into a pointer? what's the code will looks like?

Yes, you need to create an array, then copy the string at the pointer into the array. Look up strcpy() (and strncpy()).

argc and argv

argc and argv are in both C and C++. argc is a count of the command line arguments, and argv is an array of the arguments themselves. For example: foo bar bas

Code:

argc == 3;
argv[0] == foo;
argv[1] == bar;
argv[2] == bas;