（C编程）用户名和密码标识

Question

Why do I get format '%s' expects argument of type 'char*' ? How should I fix the problem?

Here are my codes:

char UserName[] = "iluvcake";
scanf("%s", &UserName);
printf("Please enter your password: \n");
char PassWord[] = "Chocolate";
scanf("%s", &PassWord);
    //if...else statement to test if the input is the correct username. 
    if (UserName == "iluvcake") 
    {
     if (PassWord == "Chocolate"){
     printf("Welcome!\n");
    }
    }else
    {
     printf("The user name or password you entered is invalid.\n");
    }

Answer 1

&UserName is a pointer to an array of char (ie, a char**). You should use

scanf( "%s", UserName );

Answer 2

#include<stdio.h>
#include<conio.h>
#include<string.h>

main(){
char name[20];
char password[10];
printf("Enter username: ");
scanf("%s",name);
printf("Enter password: ");
scanf("%s",password);
if (strcmp(name, "Admin") == 0 && strcmp(password, "pass") == 0)
printf("Access granted\n");
else printf("Access denied\n");


getch();
}

:)

Answer 3

Must be

scanf("%s", UserName);
scanf("%s", PassWord);

because UserName and PassWord are pointers to char arrays.

Answer 4

1. scanf for %s takes a char array/pointer, not pointer to it. drop the & from the scanf statements.
2. You cannot compare strings with == . Use strcmp .