將空數組傳遞給函數，在該函數中輸入值並返回值的示例

Question

我應該在main函數中創建一個空數組，然后使用兩個單獨的函數來1.接受輸入到數組中，然后2.顯示數組的值。

這是我想出的，並且在從“ int”到“ int *” [-fpermissive]無效轉換的過程中遇到轉換錯誤。 但是，我們的類要到現在才兩周才能到達指針，而這要在下周才能完成，所以我認為我們現在還不打算使用指針。

#include <iostream>
#include <iomanip>
using namespace std;

int inputFoodAmounts(int[]);
int foodFunction(int[]);


int main()
{

    int num[7];

    cout << "Enter pounds of food";
    inputFoodAmounts(num[7]);
    foodFunction(num[7]);

    return 0;

}

int inputFoodAmounts(int num[]) 
{
    for (int i = 1; i < 7; i++)
    {
        cout << "Enter pounds of food";
        cin >> num[i];
    }
}

int foodFunction(int num[])
{
    for (int j = 1; j < 7; j++)
    {   

        cout << num[j];
    }
    return 0;
}

Answer 1

您應該將num傳遞給函數； num[7]表示數組的第8個元素（超出數組的范圍），但不是數組本身。 更改為

inputFoodAmounts(num);
foodFunction(num);

---

順便說一句： for (int i = 1; i < 7; i++)看起來很奇怪，因為它只將數組從第二個元素迭代到第七個元素。

Answer 2

#include <iostream>
#include <iomanip>
using namespace std;

void inputFoodAmounts(int[]);                                           //made these two functions void you were not returning anything
void foodFunction(int[]);


int main()
{

    int num[7];

    inputFoodAmounts(num);                                              //when passing arrays just send the name 
    foodFunction(num);


    system("PAUSE");
    return 0;

}

void inputFoodAmounts(int num[])
{
    cout << "Please enter the weight of the food items: \n";            //a good practice is to always make your output readable i reorganized your outputs a bit
    for (int i = 0; i < 7; i++)                                         //careful: you wanted a size 7 array but you started index i at 1 and less than 7 so that will only give you
    {                                                                   // 1, 2, 3, 4, 5, 6 -> so only 6 
        cout << "Food "<<i +1 <<": ";
        cin >> num[i];                                                  
    }
}

void foodFunction(int num[])
{
    cout << "Here are the weight you entered: \n";
    for (int j = 0; j < 7; j++)
    {

        cout << "Food "<<j+1<<": "<<num[j]<<" pounds\n";
    }
}

我相信您會收到無效的類型錯誤，因為您正在傳遞數組num[7] 。