[英]No match for “operator<<” in std::operator
#include <iostream>
#include <string>
using namespace std;
// your code
class Dog {
public:
int age;
string name, race, voice;
Dog(int new_age,string new_name,string new_race,string new_voice);
void PrintInformation();
void Bark();
};
Dog::Dog(int new_age,string new_name,string new_race,string new_voice) {
age = new_age;
name = new_name;
race = new_race;
voice = new_voice;
}
void Dog::PrintInformation() {
cout << "Name: " << name;
cout << "\nAge: " << age;
cout << "\nRace: " << race << endl;
}
void Dog::Bark(){
cout << voice << endl;
}
int main()
{
Dog buffy(2, "Buffy", "Bulldog", "Hau!!!");
buffy.PrintInformation();
cout << "Dog says: " << buffy.Bark();
}
I'm newbie in C++ and I'm unable to figure out the error.I am getting the error at buffy.Bark(),it seems like its unable to print something which returns void. 我是C ++的新手,无法弄清错误。我在buffy.Bark()处收到错误,看来它无法打印返回void的内容。
no match for operator<< in std::operator<< >( &std::cout),((const char )
std :: operator <<>( &std :: cout),(((const char )
Either declare member function Bark
like 声明成员函数
Bark
一样
std::string Dog::Bark(){
return voice;
}
and call it like 并称它为
cout << "Dog says: " << buffy.Bark() << endl;
Or do not change the function but call it like 或不更改功能,而是调用它
cout << "Dog says: ";
buffy.Bark();
because the function has return type void. 因为该函数的返回类型为void。
Or take another dog from the dog kennel.:) 或者从狗窝里拿另一只狗。:)
Bark is defined as a void function: 树皮定义为void函数:
void Dog::Bark(){
cout << voice << endl;
}
This means that trying to do cout << buffy.Bark()
in main
is trying to cout a void
type variable, which is not possible. 这意味着试图在
main
执行cout << buffy.Bark()
尝试在cout中创建一个void
类型变量,这是不可能的。 It's likely you simply meant buffy.Bark();
您可能只是说
buffy.Bark();
, which will already output for you. ,这将已经为您输出。
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