Question

我应该在main函数中创建一个空数组，然后使用两个单独的函数来接受数组中的输入，然后2.显示数组的值。

以下是我提出的内容，并且我在从“int”到“int *”[-fpermissive]的无效转换中遇到转换错误。但是，我们的课程从现在开始两周后才开始指示，这将在下周发布，所以我假设我们还没有使用指针。

#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

您应该将num传递给函数; num[7]表示数组的第8个元素（并且它超出了数组的范围），但不是数组本身。将其更改为

inputFoodAmounts(num);
foodFunction(num);

BTW：for (int i = 1; i < 7; i++)看起来很奇怪，因为它只是将数组从第二个元素迭代到第七个元素。

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";
    }
}

我相信您收到了无效的类型错误，因为您传递了数组num[7]。

将空数组传递给函数，在该函数中输入值并返回值的示例

2 个答案: