隐式调用构造函数时std::move的行为及类型转换场景疑问

Great question—let’s break this down clearly, since there’s a common misconception about what std::move actually does, especially when implicit construction is involved.

First, let’s clear up a core misunderstanding: std::move doesn’t perform any moving on its own. All it does is take an lvalue (like the std::string objects in your vector) and cast it to an rvalue reference. This is just a hint to the compiler: “Hey, this object is no longer needed—feel free to steal its resources instead of copying them.”

Now, let’s tie this directly to your scenario with std::filesystem::path:

1. How std::filesystem::path interacts with std::string #

std::filesystem::path has two key constructors for std::string:

• A copy constructor overload that takes a const std::string&: this creates a full copy of the string’s character data.
• A move constructor overload that takes a std::string&&: this transfers ownership of the string’s internal character buffer directly to the path object, leaving the original std::string in a valid but unspecified state (usually empty, and safe to destroy afterward).

2. What happens when you use std::move during implicit construction #

When you use std::move on a std::string from your vector and pass it to something that implicitly constructs a path, the compiler will pick the most appropriate constructor:

• If you use std::move (turning the string into an rvalue), the compiler will prioritize the std::string&& constructor of path. This means no copy of the character data occurs—you’re just transferring the existing buffer to the path.
• If you skip std::move, you’re passing an lvalue, so the compiler uses the const std::string& constructor, which copies the data as you expected.

3. Applying this to your original code choice #

You mentioned you used std::copy because you thought implicit creation required it, but that’s not the case. If you don’t need the original std::vector<std::string> anymore, you can absolutely use std::move to transfer the strings to the path efficiently:

std::vector<std::string> str_paths = {"docs", "notes", "todo.txt"};
std::filesystem::path full_path;

// Standard copy (copies each string's data)
std::copy(str_paths.begin(), str_paths.end(), std::back_inserter(full_path));

// Move version (transfers string resources, no copies)
std::move(str_paths.begin(), str_paths.end(), std::back_inserter(full_path));

After the std::move call, the strings in str_paths are still safe to destroy, but their content has been moved into the full_path components—they’ll typically be empty afterward.

Key takeaway #

std::move’s behavior during implicit construction boils down to:

• It converts the source object to an rvalue reference.
• The compiler selects the constructor/function overload that accepts rvalue references (if available), triggering resource transfer instead of copying.
• If no such overload exists, it falls back to the lvalue reference (copy) overload—std::move doesn’t break anything, it just doesn’t offer efficiency gains.

In your std::filesystem::path scenario, using std::move is completely safe (and more efficient) if you don’t need the original vector anymore.

内容的提问来源于stack exchange，提问作者Jared