What is the problem in this code when I run the code it does not give the desired output

Question

#include<stdio.h>
#include<string.h>
#include<conio.h>
int main()
{
char pwd[5];
int i;
printf("Enter your password: ");
for(i=0;i<=3;i++)
{
pwd[i]=_getch();
printf("%c",pwd[i]);
 if(pwd[i]==8)
  {
    if(i>0)
    {
    printf(" \b");
    i--;
    continue;
    }
    else
        break;
  }
}
pwd[i]='\0';
if(strcmp(pwd,"abcd")==0)
{
printf("\nPassword is correct");
}
else
{
printf("\nPassword is not correct");
}
return 0;
}

I want output in such form that when the user press backspace the previous character should be deleted and the user should be allowed to re-enter the previous character but this code has some problems in it. It deletes the previous character but doesn't allow to re-enter previous corrected character instead it takes it as next character what is the problem with this code please explain?

Answer 1

There is a bug in this code. At the 18th line inside the if (i > 0) statement, you are reducing the value of variable i by 1, which is alright. But then, you are not "reinputting" pwd[i]. And remember that adding '\b' to your string doesn't remove the previous character of it. It rather "adds" a new character, but you were outsmarted by the output. So here is a working code with the bug fixed:

#include<stdio.h>
#include<string.h>
#include<conio.h>
int main()
{
    char pwd[5];
    int i;
    printf("Enter your password: ");
    for(i=0; i<=3; i++)
    {
        label:
        pwd[i]=_getch();
        printf("%c",pwd[i]);
        if(pwd[i]==8)
        {
            if(i>0)
            {
                printf(" \b");

                i--;
                goto label;
                continue;
            }
            else
                break;
        }
    }
    pwd[i]='\0';
    if(strcmp(pwd,"abcd")==0)
    {
        printf("\nPassword is correct");
    }
    else
    {
        printf("\nPassword is not correct");
    }
    return 0;
}

I think this is enough explanation from me. The rest is for you to take your time and understand the code. Happy coding!