[英]c++ - no matching function for call to
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);
#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;
}
我試圖通過引用將男孩類型的數據男孩傳遞給函數 boyInitial() 和 getBoyAges(),然后我得到了這樣的錯誤。
我很感激你的幫助!
關鍵錯誤信息如下
注意:可變大小的數組類型“long int”不是有效的模板參數
你聲明了一個變長數組(數組的大小不是一個常量表達式)
int boyNumber = 0;
cout << "how many boys do you want?" << endl;
cin >> boyNumber;
Boy boy[boyNumber];
這不是標准的 C++ 功能。
使用標准容器std::vector
代替可變長度數組。
所提供的最后一個程序有效,因為沒有可變長度數組。 數組的大小在編譯時是已知的。
至少你可以聲明函數,例如
template<class T>
void boyInitial(T *boy, size_t n ){
for( size_t i=0; i < n; i++){
std::cout << "please input boy No." << i+1 << "`s age" << std::endl;
std::cin >> boy[i].age;
}
}
該函數可以像這樣調用
boyInitial( boy, boyNumber );
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