c++ - 使用指针调用函数时出错

Question

所以我对 C++ 还是很陌生，并且已经编写了一段时间的程序。我想我正在慢慢理解它，但不断收到错误消息“Intellisense:'*' 的操作数必须是一个指针。”第 36 行第 10 列。我需要做什么来修复此错误？当我完成每个功能时，我将开始使用其他功能，对于额外的功能声明，我深表歉意

// This program will take input from the user and calculate the 
// average, median, and mode of the number of movies students see in a month.

#include <iostream>
using namespace std;

// Function prototypes
double median(int *, int);
int mode(int *, int);
int *makeArray(int);
void getMovieData(int *, int);
void selectionSort(int[], int);
double average(int *, int);

// variables
int surveyed;



int main()
{
    cout << "This program will give the average, median, and mode of the number of movies students see in a month" << endl;
    cout << "How many students were surveyed?" << endl;
    cin >> surveyed;

    int *array = new int[surveyed];

    for (int i = 0; i < surveyed; ++i)
    {
        cout << "How many movies did student " << i + 1 << " see?" << endl;
        cin >> array[i];
    }



    median(*array[surveyed], surveyed);

}


double median(int *array[], int num)
{
    if (num % 2 != 0)
    {
        int temp = ((num + 1) / 2) - 1;
        cout << "The median of the number of movies seen by the students is " << array[temp] << endl;
    }
    else
    {
        cout << "The median of the number of movies seen by the students is " << array[(num / 2) - 1] << " and " << array[num / 2] << endl;
    }

}

Answer 1

问题:

1. 表达式 *array[surveyed] 用于以下行:

   median(*array[surveyed], surveyed);
   

   是不对的。 array[surveyed] 是数组的第 surveyed 元素。它不是指针。取消引用它没有意义。

2. 声明中使用的median 的第一个参数的类型与定义中使用的类型不同。声明似乎是正确的。将实现更改为:

   double median(int *array, int num)
   

3. 修正您调用median 的方式。而不是

   median(*array[surveyed], surveyed);
   

   使用

   median(array, surveyed);