C++ 类型的非常量引用的初始化无效
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Invalid initialization of non-const reference of type
提问by jeffreyveon
In the following code, I'm not able to pass a temporary object as argument to the printAgefunction:
在以下代码中，我无法将临时对象作为参数传递给printAge函数：
struct Person {
int age;
Person(int _age): age(_age) {}
};
void printAge(Person &person) {
cout << "Age: " << person.age << endl;
}
int main () {
Person p(50);
printAge(Person(50)); // fails!
printAge(p);
return 0;
}
The error I get is:
我得到的错误是：
error: invalid initialization of non-const reference of type ‘Person&' from an rvalue of type ‘Person'
I realize that this is something to do with passing an lValue to a function expecting a rValue... Is there a way to convert my lValue to rValue by using std::move or something? I tried taking a constant parameter, but that does not seem to work.
我意识到这与将 lValue 传递给期望 rValue 的函数有关...有没有办法使用 std::move 或其他方法将我的 lValue 转换为 rValue？我尝试采用常量参数，但这似乎不起作用。
回答by Stephan Dollberg
Simply make your print function take your argument by const&. This is also logically right as it doesn't modify your argument.
只需让您的打印函数接受您的参数const&。这在逻辑上也是正确的，因为它不会修改您的论点。
void printAge(const Person &person) {
cout << "Age: " << person.age << endl;
}
The actual problem is the other way around. You are passing a temporary(rvalue) to a function which expects an lvalue.
实际的问题是相反的。您正在将临时（右值）传递给需要左值的函数。
回答by vsoftco
Or, if you have a C++11-compliant compiler, can use the so called universal reference approach, which, via reference collapsing rules, can bind to both lvalue and rvalue references:
或者，如果您有一个 C++11 兼容的编译器，可以使用所谓的通用引用方法，通过引用折叠规则，它可以绑定到左值和右值引用：
#include <iostream>
using namespace std;
struct Person {
int age;
Person(int _age): age(_age) {}
};
template<typename T> // can bind to both lvalue AND rvalue references
void printAge(T&& person) {
cout << "Age: " << person.age << endl;
}
int main () {
Person p(50);
printAge(Person(50)); // works now
printAge(p);
return 0;
}
Or, in C++14,
或者，在 C++14 中，
void printAge(auto&& person) {
cout << "Age: " << person.age << endl;
}
回答by telcom
Your code doesn't work if you run g++ or gcc compilers. You need to add constto void printAge(const Person &person). However, in Visual Studio it will work fine. I've tested for VS2010 and VS2012 and in both the following code works fine.
如果您运行 g++ 或 gcc 编译器，您的代码将不起作用。您需要添加const到void printAge(const Person &person). 但是，在 Visual Studio 中它可以正常工作。我已经对 VS2010 和 VS2012 进行了测试，并且在以下代码中都可以正常工作。
#include<iostream>
using namespace std;
struct Person {
int age;
Person(int _age): age(_age) {}
};
void printAge(Person &person) {
cout << "Age: " << person.age << endl;
}
int main () {
Person p(50);
printAge(Person(50)); // DOES NOT fail!
printAge(p);
return 0;
}
