C++11 std::move(x)函数根本没有真正 move 任何东西。它只是对 r 值的转换。为什么这样做？这不是误导吗？

最佳答案

正确的是std::move(x)只是对右值的转换 - 更具体地说是对 xvalue, as opposed to a prvalue .并且拥有一个名为 move 的 Actor 阵容也是如此。有时会迷惑人。然而，这种命名的目的不是为了混淆，而是为了使您的代码更具可读性。

历史move可追溯到 the original move proposal in 2002 .本文首先介绍右值引用，然后展示如何编写更高效的std::swap :

template <class T>
void
swap(T& a, T& b)
{
    T tmp(static_cast<T&&>(a));
    a = static_cast<T&&>(b);
    b = static_cast<T&&>(tmp);
}

I know how swap is supposed to work (copy to temporary and then exchange the values), but what is the purpose of those ugly casts?!

template <class T>
void
swap(T& a, T& b)
{
    T tmp(move(a));
    a = move(b);
    b = move(tmp);
}

Move semantics will automatically come into play when given rvalue arguments. This is perfectly safe because moving resources from an rvalue can not be noticed by the rest of the program (nobody else has a reference to the rvalue in order to detect a difference).

template <class T>
void
swap(T& a, T& b)
{
    T tmp(cast_to_rvalue(a));
    a = cast_to_rvalue(b);
    b = cast_to_rvalue(tmp);
}

But why are you casting to rvalue?

But why are you moving?

template <class T>
void
swap(T& a, T& b)
{
    T tmp(set_value_category_to_xvalue(a));
    a = set_value_category_to_xvalue(b);
    b = set_value_category_to_xvalue(tmp);
}

template <class T>
void
swap(T& a, T& b)
{
    T tmp(move(a));
    a = move(b);
    b = move(tmp);
}

template <class T>
void
swap(T& a, T& b)
{
    T tmp(static_cast<T&&>(a));
    a = static_cast<T&&>(b);
    b = static_cast<T&&>(tmp);
}

template <class T>
inline
constexpr
typename std::remove_reference<T>::type&&
set_value_category_to_xvalue(T&& t) noexcept
{
    return static_cast<typename std::remove_reference<T>::type&&>(t);
}

template <class T>
inline
constexpr
auto&&
set_value_category_to_xvalue(T&& t) noexcept
{
    return static_cast<std::remove_reference_t<T>&&>(t);
}

void
test(int& i, int& j)
{
    i = j;
}

__Z4testRiS_:                           ## @_Z4testRiS_
    .cfi_startproc
## BB#0:
    pushq   %rbp
Ltmp0:
    .cfi_def_cfa_offset 16
Ltmp1:
    .cfi_offset %rbp, -16
    movq    %rsp, %rbp
Ltmp2:
    .cfi_def_cfa_register %rbp
    movl    (%rsi), %eax
    movl    %eax, (%rdi)
    popq    %rbp
    retq
    .cfi_endproc

void
test(int& i, int& j)
{
    i = std::move(j);
}

struct X
{
    X& operator=(const X&);
};

void
test(X& i, X& j)
{
    i = j;
}

__Z4testR1XS0_:                         ## @_Z4testR1XS0_
    .cfi_startproc
## BB#0:
    pushq   %rbp
Ltmp0:
    .cfi_def_cfa_offset 16
Ltmp1:
    .cfi_offset %rbp, -16
    movq    %rsp, %rbp
Ltmp2:
    .cfi_def_cfa_register %rbp
    popq    %rbp
    jmp __ZN1XaSERKS_           ## TAILCALL
    .cfi_endproc

void
test(X& i, X& j)
{
    i = std::move(j);
}

struct X
{
    X& operator=(const X&);
    X& operator=(X&&);
};

__Z4testR1XS0_:                         ## @_Z4testR1XS0_
    .cfi_startproc
## BB#0:
    pushq   %rbp
Ltmp0:
    .cfi_def_cfa_offset 16
Ltmp1:
    .cfi_offset %rbp, -16
    movq    %rsp, %rbp
Ltmp2:
    .cfi_def_cfa_register %rbp
    popq    %rbp
    jmp __ZN1XaSEOS_            ## TAILCALL
    .cfi_endproc

关于c++ - 为什么 `std::move` 命名为 `std::move` ？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21358432/