在C++中，我该如何去使用指针来获取数组的平均值？

Question

我是编程新手，我仍然遇到数组，指针和函数的麻烦。我想知道这有什么问题，以及我如何解决这个问题。具体为什么指针不能与该函数一起工作。这里是我试图编写的程序：编写一个程序，DYNAMICALLY创建一个指向数组足够大的数组的指针，以保存用户定义的测试分数。一旦输入了所有分数（在主函数中），该数组就应该被传递给一个返回平均分数的DOUBLE的函数。在用户输出中，平均得分应格式化为两位小数。使用指针表示法;不要使用数组表示法。

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

double getAverage(int, int); 

int main() 
{ 
    int size = 0; 
    cout << "How many scores will you enter? "; 
cin >> size; 

unique_ptr<int[]> ptr(new int[size]); 


cout << endl; 
int count = 0; 

//gets the test scores 

for (count = 0; count < size; count++) 
{ 
    cout << "Enter the score for test " << (count + 1) << ": "; 
    cin >> ptr[count]; 
    cout << endl; 
} 
//display test scores 
cout << "The scores you entered are:"; 
for (count = 0; count < size; count++) 
    cout << " " << ptr[count]; 
cout << endl; 

double avg; 
avg = getAverage(ptr, size); 
cout << setprecision(2) << fixed << showpoint << endl; 

cout << "The average is " << avg << endl; 

return 0; 
} 

double getAverage(int *ptr, int size) 
{ 
double average1; 
double total = 0; 
for (int count = 0; count < size; count++) 
{ 
    total = total + *(ptr + count); 
} 
average1 = total/size; 

return average1; 
} 

Answer 1

首先，您的功能getAverage()与您定义的原型不同。其次，您尝试将std::unique_ptr<int []>对象转换为函数，而不是期望int*。但是std::unique_ptr<int []>是与int*不同的类型，不能隐式转换。 所以要通过int *使用std::unique_ptr::get函数。 像

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

double getAverage(int *, int); 

int main() 
{ 
    int size = 0; 
    cout << "How many scores will you enter? "; 
cin >> size; 

unique_ptr<int[]> ptr(new int[size]); 


cout << endl; 
int count = 0; 

//gets the test scores 

for (count = 0; count < size; count++) 
{ 
    cout << "Enter the score for test " << (count + 1) << ": "; 
    cin >> ptr[count]; 
    cout << endl; 
} 
//display test scores 
cout << "The scores you entered are:"; 
for (count = 0; count < size; count++) 
    cout << " " << ptr[count]; 
cout << endl; 

double avg; 
avg = getAverage(ptr.get(), size); 
cout << setprecision(2) << fixed << showpoint << endl; 

cout << "The average is " << avg << endl; 

return 0; 
} 

double getAverage(int *ptr, int size) 
{ 
double average1; 
double total = 0; 
for (int count = 0; count < size; count++) 
{ 
    total = total + *(ptr + count); 
} 
average1 = total/size; 

return average1; 
} 

Answer 2

你已经差不多了。只需要改变这个块：

using namespace std; 

double getAverage(int*, int); // <--here 

int main() 
{ 
    int size = 0; 
    cout << "How many scores will you enter? "; 
    cin >> size; 
    int *ptr = new int[size]; //<--and here (but you'll need to delete it later) 
    //unique_ptr<int[]> ptr(new int[size]); 

我不建议混合智能指针和标准指针，直到你有一个更好的理解标准指针。编辑：我的意思是混合它们的相同的有效指针。

Answer 3

在分配有书面

使用指针符号;不要使用数组符号

这意味着你不应该使用下标。

主要变量ptr被声明为具有类型std::unique_ptr<int[]>。

unique_ptr<int[]> ptr(new int[size]); 

您正在试图将它传递给函数getAverage

avg = getAverage(ptr, size); 

具有类型int的相应参数。

double getAverage(int, int); 

虽然那么你定义的函数具有在任何情况下的类型int *的参数存在从类型std::unique_ptr<int[]>到类型int *没有显式转换。您应该使用智能指针的方法get。

此外，函数参数应声明为const int *，因为数组在函数中没有更改。

程序可以看看下面的方式

#include <iostream> 
#include <iomanip> 
#include <memory> 

double getAverage(const int *a, size_t n) 
{ 
    double total = 0.0; 

    for (const int *p = a; p != a + n; ++p) total += *p; 

    return n == 0 ? total : total/n; 
} 

int main() 
{ 
    size_t n = 0; 

    std::cout << "How many scores will you enter? "; 
    std::cin >> n; 

    std::unique_ptr<int[]> ptr(new int[n]); 

    std::cout << std::endl; 

    size_t i = 0; 
    for (int *current = ptr.get(); current != ptr.get() + n; ++current) 
    { 
     std::cout << "Enter the score for test " << ++i << ": "; 
     std::cin >> *current; 
    } 

    std::cout << std::endl; 

    std::cout << "The scores you entered are:"; 
    for (const int *current = ptr.get(); current != ptr.get() + n; ++current) 
    { 
     std::cout << " " << *current; 
    } 
    std::cout << std::endl; 

    double avg = getAverage(ptr.get(), n); 

    std::cout << std::setprecision(2) << std::fixed << std::showpoint; 

    std::cout << "\nThe average is " << avg << std::endl; 

    return 0; 
} 

程序输出可能看起来像

How many scores will you enter? 10 

Enter the score for test 1: 1 
Enter the score for test 2: 2 
Enter the score for test 3: 3 
Enter the score for test 4: 4 
Enter the score for test 5: 5 
Enter the score for test 6: 6 
Enter the score for test 7: 7 
Enter the score for test 8: 8 
Enter the score for test 9: 9 
Enter the score for test 10: 10 

The scores you entered are: 1 2 3 4 5 6 7 8 9 10 

The average is 5.50