警告：初始化使 integer 来自指针而无需强制转换，即使我输入正确的用户和密码，output 也只显示其他部分

Question

#include<stdio.h>
int main(){
    char user = "welcome";
    printf("Enter UserName \n");
    scanf("%s",&user);

    char pass = "hi";
    printf("Enter Password: \n");
    scanf("%s",&pass);

    if(user=="welcome" && pass == "hi"){
            printf("Login Successful");
    }
    else{
        printf("Retry");
    }
    return 0;
}

This program asks to enter username and password then checks if entered correctly and prints suitable statement.

However, for me it only shows else part even if the data entered is correct.

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Answer 1

The program is entirely wrong.

In these declarations

char user = "welcome";

and

char pass = "hi";

you declared objects of the type char that you are trying to initialize by expressions of the type char * (the string literals used as initializers are implicitly converted to pointers to their first elements).

Instead you need to declare character arrays where you will read the user name and his password.

For example

char user[10];
char pass[10];

The calls of scanf can look the following way

scanf("%9s", user);
scanf("%9s", pass);

In this if statement

if(user=="welcome" && pass == "hi"){

you are comparing pointers. Instead you need to compare strings like

if( strcmp( user, "welcome" ) == 0 && strcmp( pass, "hi" ) == 0 ){

To use the function strcmp you need to include header <string.h>

#include <string.h>

Answer 2

Declare the username and password fields something like this:

char user[1024], pass[1024];

Currently they can only hold a single character. You can change the 1024 to the maximum number of characters that you'd assume the user would input. You should also keep in mind strings cannot be compared with user=="welcome" ! Use the strcmp function inside string.h instead.