In the following code, I'm not able to pass a temporary object as argument to the printAge
function:
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:
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.
Simply make your print function take your argument by const&
. This is also logically right as it doesn't modify your argument.
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.
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:
#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,
void printAge(auto&& person) {
cout << "Age: " << person.age << endl;
}
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