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

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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?

• c++
• rvalue
• lvalue

Share Improve this question Follow asked Nov 18, 2024 at 11:21 AikAik 6555 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

Sorted by: 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}));

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?

• c++
• rvalue
• lvalue

Share Improve this question Follow asked Nov 18, 2024 at 11:21 AikAik 6555 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

Sorted by: 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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