使用指针调用函数时出现错误

Question

所以即时通讯对C++来说仍然很新颖，并且现在已经在执行一段程序。我认为我正在慢慢获得它，但不断收到错误“智能感知：'*'的操作数必须是指针。”在第36行第10列。我需要做些什么来解决这个错误？我就要去其他的功能，因为我完成了额外的函数声明每一个难过

// 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. 在下面的行中使用的表达*array[surveyed]：

   median(*array[surveyed], surveyed); 
   

   是不正确的。 array[surveyed]是数组中的surveyed个元素。它不是一个指针。解引用它是没有意义的。

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

   double median(int *array, int num) 
   

3. 解决您拨打median的方式。取而代之的

   median(*array[surveyed], surveyed); 
   

   使用

   median(array, surveyed);