Help with very, very basic JAVA task

[FMQ203]

So, first, I want to make it clear, I'm not into programming, I've never been, but I'm taking a course about JAVA, and we got a task assigned about making a little game.

Basically, the program makes a 5 digits random number (every digit is different) and the user has to guess it, and, after every attempt the program prints out the amount of digits that are right in the correct spot, and the amount that its right but not in the correct spot. For the the first ones I make this and it works! (How to look for them after the user enters a number)(It's in Spanish because I'm from Uruguay ) :

 

public static int cantBien (String ingreso, String numero) {

        int contador = 0;

        for (int i = 0; i < largo-1; i++) {

           if (ingreso.charAt(i) == numero.charAt(i)) {

                contador++;

           } 

        }

        return contador;

 

But I'm struggling to get the second ones work, I tried this but its returns is wrong data:

 

   public static int cantReg (String ingreso, String numero) {

        int contador = 0;

        for (int j=0; j < largo-1; j++) {

            for (int k=0; k < largo-1; k++) {

                if(ingreso.charAt(k) == numero.charAt(k)) {

                    contador++;

                }

            }

        }

        return (contador-cantBien(ingreso,numero));

 

Hope you can help me!

 

[Nineshadow]

@FMQ203

 

Use the code tags.

 

Anyway, it seems the function, oh wait, it's method in Java, that returns the number of wrong elements at the same possition doesn't work correctly, am I right?

Lemme make a quick one myself. I don't remember a whole lot of java, but I tested this and it works.

public static int wrongCount(String original, String guess){int k = 0;for(int i = 0; i < original.length(); i++)if(original.charAt(i)!=guess.charAt(i)) k++;return k;}

[log234]

I think he meant that it should return the amount of characters that are correct, but in the wrong spot.

Here is a simple method for that 

public static int wrongCount(String original, String guess) {    int count = 0;    for (int i = 0; i < original.length(); i++) {        for (int j = 0; j < guess.length(); j++) {            if (original.charAt(i) == guess.charAt(j) && i != j) count++;        }    }    return count;}

This method is quite simple, it checks for any correct guesses, and then it checks if the correct guess is in a different spot as the original. If both are true, then add 1 to the final count.

 

Hope this helps! 

[ZuitAMB]

 public static int cantReg (String ingreso, String numero) {        int contador = 0;        for (int j=0; j < largo-1; j++) {            for (int k=0; k < largo-1; k++) {                if(ingreso.charAt(k) == numero.charAt(k)) {//check this line                    contador++;                }            }        }        return (contador-cantBien(ingreso,numero));}

simple swapping a k for a j will do the trick

 

[log234]

 

simple swapping a k for a j will do the trick 

 

 

Ah, yes, if the goal is to use the first method to solve the last method, then yes, this is the answer! 

[FMQ203]

 

 public static int cantReg (String ingreso, String numero) {        int contador = 0;        for (int j=0; j < largo-1; j++) {            for (int k=0; k < largo-1; k++) {                if(ingreso.charAt(k) == numero.charAt(k)) {//check this line                    contador++;                }            }        }        return (contador-cantBien(ingreso,numero));}

simple swapping a k for a j will do the trick

 

 

I used this, thanks!!

[Nineshadow]

I think he meant that it should return the amount of characters that are correct, but in the wrong spot.

Aaah.Didn't quite understand that.