創建一個方法來返回數組的值的平均值

Question

我試圖在傳遞一個數組對象數組的類中創建一個靜態方法，並返回數組中的對象的平均值。

public static double calcAverage() { 
    int sum = 0; 
    for (int i=0; i < people.length; i++) 
      sum = sum + people[i]; 
    double calcAverage() = sum/people.length 
    System.out.println(people.calcAverage()); 
} 

該代碼是越來越編譯錯誤，但我正朝着正確的方向？

Answer 1

public static double calcAverage() { 
    int sum = 0; 
    for (int i=0; i < people.length; i++) 
      sum = sum + people[i]; 
    double calcAverage() = sum/people.length 
    System.out.println(people.calcAverage()); 
} 

變化

double calcAverage() = sum/people.length 

到

double average = sum/(double)people.length; 

（聲明一個新變量的正確方法）

變化

 System.out.println(people.calcAverage()); 

到

return average; 

（如果你想打印調用該函數的結果，你應該總是做它的功能，例如外做到在main調用該函數和存儲返回的結果）

後，所以我們有：

public static double calcAverage() { 
    int sum = 0; 
    for (int i=0; i < people.length; i++) 
    { 
     sum = sum + people[i]; 
    } 
    double average = sum/(double)people.length; 
    return average; 
} 

Answer 2

你的親密。雖然我看到一些錯誤。

首先你的總和= sum + people [i];

people [i] returns object is not an integer，so add a object to a integer wont work。

秒，你在calcAverage方法內調用calcAverage（），這可能不是你想要做的。這樣做被稱爲遞歸，但我認爲你應該在calcAverage（）之外調用方法。

Answer 3

// pass people as a parameter 
public static double calcAverage(int[] people) { 
    // IMPORTANT: this must be a double, otherwise you're dividing an integer by an integer and you will get the wrong answer 
    double sum = 0; 
    for (int i=0; i < people.length; i++) { 
     sum = sum + people[i]; 
    } 
    // remove the() 
    double result = sum/people.length; 
    System.out.println(result); 

    // return something 
    return result; 
} 


// example 
int[] myPeople = {1,2,3,4,5,6,7,8}; 
double myPeopleAverage = calcAverage(myPeople);