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");
}
5 Answers
&UserName is a pointer to an array of char (i.e., a char**). You should use
scanf( "%s", UserName );
#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();
}
:)
- scanf for %s takes a char array/pointer, not pointer to it. drop the
&from thescanfstatements. - You cannot compare strings with
==. Usestrcmp.
Must be
scanf("%s", UserName);
scanf("%s", PassWord);
because UserName and PassWord are pointers to char arrays.
#include <stdio.h>
int main(){
int serial;
int code;
printf("Type the Serial Here: ");
scanf("%d", &serial);
if(serial== 1590){
printf("Almost there, now type the code: ");
scanf("%d", &code);
if(code==2359)
{
printf("\nYOU ARE DONE, ENJOY");
}
else{
printf("TRY AGAIN NEXT TIME");
}
}
else{
printf("OPPS");
}
