Question

我已经尝试在其他线程中在这里寻找答案，但是我似乎仍然找不到答案。我收到一个错误：应该在private static int getUserMHR（age）行。这是我的代码：

import java.util.Scanner;

public class Assignment2
{
//class variables
//a scanner object to read input from the keyboard - can be used by any method
static Scanner input = new Scanner(System.in);

public static void main(String[] args)
{
  String name; // the user's name
  int age; // the user's age
  String gender; // the user's gender
  float height; // the users height in inches
  float weight; // the users weight in pounds
  int mhr; //the user's maximum heart rate
  int thr; //the user's target heart rate
  float bmi; //the user's body mass index

  //display the purpose of the program
  displayPurpose();
  //get the user's name
  name = getUserName();
  //get the user's age
  age = getUserAge();
  //get the user's gender
  gender = getUserGender();
  //get the user's height
  height = getUserHeight();
  //get the user's weight
  weight = getUserWeight();
  //get the user's maximum heart private
  mhr = getUserMHR();

   System.out.println(mhr);

}// of main

private static void displayPurpose()
{
    System.out.println("****************************************************************************************************");
    System.out.println("            This application takes the user's name, age, gender, height and weight");
    System.out.println("It then display's the user's maximum heart rate, target heart rate, BMI (Body Mass Index), and their");
    System.out.println("                                              weight class");
    System.out.println("****************************************************************************************************");
}//displayPurpose

private static String getUserName()
{
  //local variable
  String username;

  //ask for user's name
  System.out.println("Please input your name: ");
  //input the user's name
  username = input.nextLine();

  //return user's name
  return username;

}//getUserName();

private static int getUserAge()
{
  //local variable
  int userage;

  //ask for user's age
  System.out.println("Please input your age: ");
  //input the user's age
  userage = input.nextByte();

  //return user's age
  return userage;
}//getUserAge();

private static String getUserGender()
{
  //local variable
  String gender;

  //ask for user's gender
  System.out.println("Please input your gender (M or F): ");
  //input the user's gender
  gender = input.nextLine();

  //return user's gender
  return gender;
}//getUserGender();

private static float getUserHeight()
{
  //local variable
  float height;

  //ask for user's height
  System.out.println("Please input your height (in inches): ");
  //input the user's height
  height = input.nextFloat();

  //return user's height
  return height;
}//getUserHeight();

private static float getUserWeight()
{
  //local variable
  float weight;

  //ask for user's weight
  System.out.println("Please input your weight (in pounds): ");
  //input the user's weights
  weight = input.nextFloat();

  //return user's weight
  return weight;
}//getUserWeight();

private static int getUserMHR(age)
{
  int mhrcalc;
  mhrcalc = (age - 220); //calculate the user's maximum heart private

  return mhrcalc;
}

}

我尝试在年龄之前放置一个int，并从实际方法中删除该int，但似乎没有任何效果。感谢您的帮助。

Answer 1

您的方法getUserMHR具有单个参数age。但是在Java中，您必须指定参数的类型。由于age很可能是整数，因此您只能使用基本类型int。

private static int getUserMHR( int age ) { //... }

Answer 2

您需要指定age是什么类型，这里是int：

private static int getUserMHR(int age){
  int mhrcalc;
  mhrcalc = (age - 220); //calculate the user's maximum heart private
  return mhrcalc;
}

当您要求一个参数时，在调用该方法时需要给它

mhr = getUserMHR(age);

此外，您不需要拆分声明和赋值的每一句（这里似乎您希望对评论进行规范，这样就可以了） int t; t=5; ==> int t=5;

方法：预期的标识符

2 个答案: