c++ - Why does passing rvalue to lvalue reference work in this case? - Stack Overflow

I was doing Leetcode when I stumbled upon this interesting technique. Here it is in short:#include <

I was doing Leetcode when I stumbled upon this interesting technique. Here it is in short:

#include <iostream>
#include <vector>

void print(std::vector<int>& vec)
{
    for (int i : vec)
    {
        std::cout << i << ", ";
    }
}

int main()
{
    print(std::vector<int>(5) = {1,2,3,4,5});

    return 0;
}

We are passing seemingly rvalue to an lvalue reference, which is usually illegal, but in this specific case, it works somehow. Is there any explanation or maybe I'm missing something?

I was doing Leetcode when I stumbled upon this interesting technique. Here it is in short:

#include <iostream>
#include <vector>

void print(std::vector<int>& vec)
{
    for (int i : vec)
    {
        std::cout << i << ", ";
    }
}

int main()
{
    print(std::vector<int>(5) = {1,2,3,4,5});

    return 0;
}

We are passing seemingly rvalue to an lvalue reference, which is usually illegal, but in this specific case, it works somehow. Is there any explanation or maybe I'm missing something?

Share Improve this question asked Nov 18, 2024 at 11:21 AikAik 655 bronze badges 4
  • The argument still has a valid life cycle (for the duration of the call) and thus can be converted to a reference – Pepijn Kramer Commented Nov 18, 2024 at 11:27
  • This is why I consider it a good idea to lvalue-ref-qualify most assignment operators. std::vector<int>(5) = {1,2,3,4,5} itself should be flagged – StoryTeller - Unslander Monica Commented Nov 18, 2024 at 11:32
  • The left-hand 5 is pretty pointless. – user17732522 Commented Nov 18, 2024 at 11:33
  • Related stackoverflow/questions/61141894/… – cigien Commented Nov 18, 2024 at 11:55
Add a comment  | 

1 Answer 1

Reset to default 5

operator= of std::vector returns an lvalue reference and is callable on rvalue as well as lvalue object expressions (because it is just a normal non-static member function without ref-qualifier).

So std::vector<int>(5) = {1,2,3,4,5} is permitted and is an lvalue expression that can be bound by the lvalue reference in the parameter.

Of course, that's terrible style. print should just have a const lvalue reference parameter and then

print({1,2,3,4,5});

would simply work.

And even if one really wants to pass a rvalue to a non-const lvalue reference, it would be better to use a function specifically for that which clearly communicates the intent, e.g.:

template<typename T>
T& as_lvalue(T&& t) { return static_cast<T&>(t); }

//...

print(as_lvalue(std::vector<int>{1,2,3,4,5}));

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