向函数传递二维数组

C++ Passing 2d arrays to functions

本文关键字:二维数组 函数      更新时间:2023-10-16

要理解计算数组的平均值,我们需要在下面声明一个函数。逐行循环读取

double getAverage(int arr[], int size)
{
  int    i, sum = 0;       
  double avg;          
  for (i = 0; i < size; ++i){
    sum += arr[i];
   }
  avg = double(sum) / size;    
  return avg;
}

之后,将该值调用给main函数。

#include <iostream>
using namespace std;
// function declaration:
double getAverage(int arr[], int size);
int main ()
{
   // an int array with 5 elements.
   int balance[5] = {1000, 2, 3, 17, 50};
   double avg;
   // pass pointer to the array as an argument.
   avg = getAverage( balance, 5 ) ;
   // output the returned value 
   cout << "Average value is: " << avg << endl; 
   return 0;
}

我的问题是,如果我想计算row*col的平均值呢?我要申报这样的东西吗?设row和col的大小为arr[3][6]

double getAverage(int arr[][6], int noOfrows, int noOfcol)
{
    float sum=0, average;
    for (int i = 0 ; i < noOfrows ; i++) {
        for (int j = 0; j<noOfcol; j++) {    
        sum = sum + arr[i][j];    
        }
    }
        average = (float)sum / (float)(noOfcol*noOfrows);
        cout << "   " << average;
        return average; 
}

这是我的代码

int main()
{   
    int sales[3][6] = {{1000, 800, 780, 450, 600, 1200},
                       {800, 900, 500, 760, 890, 1000},
                        {450, 560, 570, 890, 600, 1100}};
    int avg;
    int choice;//menu choice
    const int computeAverage_choice = 1,
              computeTotal_choice = 2,
              listMaxMin_choice = 3,
              Exit_choice = 4;
    do
    {   
            //displayMenu(); // Show Welcome screen
            choice = displayMenu();
            while (choice < 1 || choice > 4)
         {
               cout << "Please enter a valid menu choice: " ;
               cin >> choice;
         }
            //If user does not want to quit, proceed.
         if (choice != Exit_choice)
         {  
                    switch (choice)
                    {
                           case computeAverage_choice:
                                avg = computeAverage(sales, 3, 6);
                                cout<<"The averge:" << avg;
                                break;
                           case computeTotal_choice:
                                //reserves
                                break;
                           case listMaxMin_choice:
                                //reserves
                                break;
                    }
         }
         } while (choice != Exit_choice);
            return 0;
}

如果你的函数是这样声明的:double getAverage(int arr[][6], int noOfrows, int noOfcol) 但是你试图用avg = getAverage( balance, 5 ) ;[只有2个参数]来调用它,你的编译器应该返回一个错误。

请将呼叫调整为avg = getAverage( balance, 5 */num or rows*/ , 6 */num of cols*/) ;

#include <iostream>
using namespace std;
 double getAverage(int arr[][6], int noOfrows, int noOfcol)
{
    float sum=0, average;
    for (int i = 0 ; i < noOfrows ; i++) {
        for (int j = 0; j<noOfcol; j++) {
        sum = sum + arr[i][j];
        }
    }
        average = (float)sum / (float)(noOfcol*noOfrows);
        cout << "   " << average;
        return average; 
}

 int main()
 {
    int a[2][6] = {{1,2,3,4,5,6},{2,3,4,5,6,7}};
    getAverage(a, 2, 6);   // OK
    getAverage( a, 5 ) ;   // compile error
    return 0;
 }

"我要声明这样的东西吗?"

是的。这就是如何声明一个函数,该函数接受包含6个int的数组的数组。

如果大小未知,有两种解决方法:使用std::vector(或者对于数组的数组,使用向量的向量)。或者您可以使用模板为数组的行和列:

template<size_t noOfrows, size_t noOfcol>
int getAverage(int const (&arr)[noOfrows][noOfcol])
{
    ...
}

这样你只需要以数组作为参数调用它,编译器将推断出行数和列数:

int balance1[4][7] = { ... };
int average1 = getAverage(balance1);
int balance2[2][5] = { ... };
int average2 = getAverage(balance2);

无论数组或子数组的大小,

我不太明白你的问题。如果您担心声明行和列并将其传递给函数,您可以通过将数组作为参数传递或将指向数组的指针作为参数传递来实现。我已经在#define中定义了行和色的长度。

定义
#include <iostream>
using namespace std;
#define NO_OF_ROWS 2
#define NO_OF_COLS 3

主要功能
int main ()
{
// an int array with 5 elements
int balance[5] = {1000, 2, 3, 17, 50};
int ar[2][3] = {{5,5,5},{10,10,10}};
double avg, avgRC;
// pass pointer to the array as an argument.
avg = getAverage( balance, 5 ) ;
avgRC = getAvg(ar);
// output the returned value 
cout << "Average value is: " << avg << endl; 
cout << "Average of row*col: " << avgRC << endl;
return 0;
}

返回row*col平均值的GetAvg函数

double getAvg(int arr[][NO_OF_COLS])
{
 float sum=0, average;
   for (int i = 0 ; i < NO_OF_ROWS ; i++) {
    for (int j = 0; j<NO_OF_COLS; j++) {
    sum = sum + arr[i][j];
    }
 }
    average = (float)sum / (float)(NO_OF_ROWS*NO_OF_COLS);
    cout << "   " << average;
    return average; 
}

由于int sales[3][6]int sales[3 * 6]具有相同的布局,您可以重用以前的代码并简单地调用

const double avg = getAverage(&sales[0][0], 3 * 6);

如果你想传递数组,最好的方法是通过引用传递(使用非直观的语法),并让模板推断大小:

template <std::size_t R, std::size_t  C>
double getAverage(const int (&a)[R][C]) {
    return std::accumulate(&a[0][0], &a[0][0] + R * C, 0.) / (R * C);
}