How do I verify this equation generator with three operands and two operators? Each time I run it it is always wrong even when I input the correct answer.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Random {
public static void main(String[] args) {
System.out.println("Difficulty Random Generation Algorithms:");
easyDifficulty();
}
public static void easyDifficulty() {
int operand1, operand2, operand3, operator1, operator2;
operand1 = (int) (Math.random() * 25) + 1; // questions will only contain operands up to the value of 25.
operand2 = (int) (Math.random() * 25) + 1; //"
operand3 = (int) (Math.random() * 25) + 1; //"
operator1 = (int) (Math.random() * 4) + 1; //Possibility of four operators: +, -, * and /.
operator2 = (int) (Math.random() * 4) + 1; //"
String operation1 = null; //String initially null value.
int actualAnswer = 0; //int initially zero value.
switch (operator1) {
case 1:
operation1 = "+";
actualAnswer = operand1 + operand2;
break;
case 2:
operation1 = "-";
actualAnswer = operand1 - operand2;
break;
case 3:
operation1 = "*";
actualAnswer = operand1 * operand2;
break;
case 4:
operation1 = "/";
actualAnswer = operand1 / operand2;
break;
}
String operation2 = null; //String initially null value.
switch (operator2) {
case 1:
operation2 = "+";
actualAnswer = operand1 + operand2;
break;
case 2:
operation2 = "-";
actualAnswer = operand1 - operand2;
break;
case 3:
operation2 = "*";
actualAnswer = operand1 * operand2;
break;
case 4:
operation2 = "/";
actualAnswer = operand1 / operand2;
break;
}
System.out.println("Question: " + operand1 + operation1 + operand2 + operation2 + operand3);
String yourGuess = null;
try {
System.out.print("Your guess: ");
yourGuess = new BufferedReader(new InputStreamReader(System.in)).readLine();
} catch (IOException e) {
e.printStackTrace();
}
if (actualAnswer == Integer.parseInt(yourGuess)) {
System.out.println("Correct!");
} else {
System.out.println("\nIncorrect!");
System.out.println("Correct answer: " + actualAnswer);
}
}
}
Aucun commentaire:
Enregistrer un commentaire