int main() {
int x;
auto& getx = [&]() {
return x;
}
getx() = 1;
}
This doesn't seem possible, as it gives me:
error: cannot bind non-const lvalue reference of type 'main()::<lambda()>&' to an rvalue of type 'main()::<lambda()>'|
Why ? and how to do it
A lambda expression creates a temporary object (called a closure) of an unknown class type. Temporary objects can't be assigned to non-const references. That means you need
auto&& getx = [&]() {
return x;
}
so that you get an rvalue reference to the closure, or
auto getx = [&]() {
return x;
}
so you just get the closure.
That will get the code to compile, but still needs one more bit to make getx's return value to be a reference to x. To do that you need to add
auto getx = [&]() -> int& {
return x;
}
// or
auto getx = [&]() -> auto& {
return x;
}
// or
auto getx = [&]() -> decltype(auto) {
return x;
};
Also note that main must always return an int. You should turn up your compiler warning level so that it errors out if you try and use void main.
getx to be? Do you want it to be a reference to x, or do you want it to return a reference to x, or do you want to to return the value of x?getx() = 1;You have to specify that lambda returns reference to int - int&, and call closure by ():
auto& getx = [&]() -> int& { // -> int& added
return x;
}(); // () added
Temporary value cannot be bound to lvalue reference. If you want to make getx to be as a reference to x variable, you have to return reference from your lambda.
