stdmove | 易学教程

Why to use std::move despite the parameter is an r-value reference

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问题 I am confused about using std::move() in below code: If I uncomment line at (2) the output would be: 1 2 3 but if I uncomment line at (1) output would be nothing which means that move constructor of std::vector was called! Why do we have to make another call to std::move at (1) to make move constructor of std::vector to be called? What I understood that std::move get the r-value of its parameter so, why we have to get the r-value of r-value at (1)? I think this line _v = rv; at (2) is more

Why to use std::move despite the parameter is an r-value reference

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Why to use std::move despite the parameter is an r-value reference

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How does std::move invalidates the value of original variable?

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问题 In the following examples from cpp reference: #include <iostream> #include <utility> #include <vector> #include <string> int main() { std::string str = "Hello"; std::vector<std::string> v; // uses the push_back(const T&) overload, which means // we'll incur the cost of copying str v.push_back(str); std::cout << "After copy, str is \"" << str << "\"\n"; // uses the rvalue reference push_back(T&&) overload, // which means no strings will be copied; instead, the // Contents of str will be moved

Why doesn't C++ move construct rvalue references by default? [duplicate]

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This question already has an answer here: Rvalue Reference is Treated as an Lvalue? 4 answers Lvalue reference constructor is called instead of rvalue reference constructor 1 answer Say I have the following function void doWork(Widget && param) // param is an LVALUE of RRef type { Widget store = std::move(param); } Why do I need to cast param back to an rvalue with std::move() ? Shouldn't it be obvious that the type of param is rvalue since it was declared in the function signature as an rvalue reference? Shouldn't the move constructor be automatically invoked here on this principle alone? Why

Should I return an rvalue reference (by std::move'ing)?

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A C++Next blog post said that A compute(…) { A v; … return v; } If A has an accessible copy or move constructor, the compiler may choose to elide the copy. Otherwise, if A has a move constructor, v is moved. Otherwise, if A has a copy constructor, v is copied. Otherwise, a compile time error is emitted. I thought I should always return the value without std::move because the compiler would be able to figure out the best choice for users. But in another example from the blog post Matrix operator+(Matrix&& temp, Matrix&& y) { temp += y; return std::move(temp); } Here the std::move is necessary

Should I return an rvalue reference (by std::move'ing)?

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问题 A C++Next blog post said that A compute(…) { A v; … return v; } If A has an accessible copy or move constructor, the compiler may choose to elide the copy. Otherwise, if A has a move constructor, v is moved. Otherwise, if A has a copy constructor, v is copied. Otherwise, a compile time error is emitted. I thought I should always return the value without std::move because the compiler would be able to figure out the best choice for users. But in another example from the blog post Matrix

Why doesn't C++ move construct rvalue references by default? [duplicate]

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问题 This question already has an answer here: Rvalue Reference is Treated as an Lvalue? 4 answers Lvalue reference constructor is called instead of rvalue reference constructor 1 answer Say I have the following function void doWork(Widget && param) // param is an LVALUE of RRef type { Widget store = std::move(param); } Why do I need to cast param back to an rvalue with std::move() ? Shouldn't it be obvious that the type of param is rvalue since it was declared in the function signature as an

What is std::move(), and when should it be used?

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问题 What is it? What does it do? When should it be used? Good links are appreciated. 回答1: Wikipedia Page on C++11 R-value references and move constructors In C++11, in addition to copy constructors, objects can have move constructors. (And in addition to copy assignment operators, they have move assignment operators.) The move constructor is used instead of the copy constructor, if the object has type "rvalue-reference" ( Type && ). std::move() is a cast that produces an rvalue-reference to an