我的终端消息

zo@laptop:~/Desktop$ g++ stack.cpp 
stack.cpp: In function ‘int main(int, char**)’:
stack.cpp:50:19: error: no matching function for call to ‘boyInitial(Boy [boyNumber])’
     boyInitial(boy);
                   ^
stack.cpp:17:6: note: candidate: template<class T, int N> void boyInitial(T (&)[N])
 void boyInitial(T (&boy)[N]){
      ^~~~~~~~~~
stack.cpp:17:6: note:   template argument deduction/substitution failed:
stack.cpp:50:19: note:   variable-sized array type ‘long int’ is not a valid template argument
     boyInitial(boy);
                   ^
stack.cpp:51:19: error: no matching function for call to ‘getBoyAges(Boy [boyNumber])’
     getBoyAges(boy);
                   ^
stack.cpp:34:6: note: candidate: template<class T, int N> void getBoyAges(T (&)[N])
 void getBoyAges(T (&boy)[N]){
      ^~~~~~~~~~
stack.cpp:34:6: note:   template argument deduction/substitution failed:
stack.cpp:51:19: note:   variable-sized array type ‘long int’ is not a valid template argument
     getBoyAges(boy);

我的程序

#include <iostream>

class People{
    public:
        char *name;
        int age;
};

class Boy : public People{
    public:
        void say(void){
            std::cout << "i`m the most handsome!" << std::endl;
        }
};

template<class T, int N> 
void boyInitial(T (&boy)[N]){
    int i;
    for(i=0;i<N;i++){
        std::cout << "please input boy No." << i+1 << "`s age" << std::endl;
        std::cin >> boy[i].age; 
    }
}

template<class T, int N>
void getBoyAges(T (&boy)[N]){
    int i;
    for(i=0;i<N;i++){
        std::cout << boy.age << std::endl;
    }
}

int main(int argc, char **argv){
    using namespace std;
    int boyNumber = 0;
    cout << "how many boys do you want?" << endl;
    cin >> boyNumber;
    Boy boy[boyNumber];
    boyInitial(boy);
    getBoyAges(boy);
    return 0;
}

描述

我试图通过引用将男孩类型的数据男孩传递给函数 boyInitial() 和 getBoyAges()，然后我得到了这样的错误。

我试图模仿这里的代码，然后得到错误。

我很感激你的帮助！