I'm writing an implementation of standard c++ library for study.
The C++11 standard says that for_each
returns std::move(f)
.
template <class InputIterator, class Function>
Function for_each(InputIterator first, InputIterator last, Function f);
Returns: std::move(f).
I thought that function scope local variable is move-constructed when it's returned.
Should I return move(f)
explicitly?
Yes, unless you catch it, or unless you mark the destructor noexcept(false) . And in the latter case you have to be pretty careful or it may result in std::terminate() anyway.
std::move is used to indicate that an object t may be "moved from", i.e. allowing the efficient transfer of resources from t to another object. In particular, std::move produces an xvalue expression that identifies its argument t . It is exactly equivalent to a static_cast to an rvalue reference type.
applies a function object to the first n elements of a sequence. (function template) ranges::for_each. (C++20) applies a function to a range of elements.
From Josuttis 's The C++ Standard Library
You don’t have to and should not move() return values. According to the language rules, the standard specifies that for the following code
X foo ()
{
X x;
...
return x;
}
the following behavior is guaranteed:
• If X has an accessible copy or move constructor, the compiler may choose to elide the copy. This is the so-called (named) return value optimization ((N)RVO), which was specified even before C++11 and is supported by most compilers.
• Otherwise, if X has a move constructor, x is moved.
• Otherwise, if X has a copy constructor, x is copied.
• Otherwise, a compile-time error is emitted.
From §25.2.4 (for_each)
Requires:Function shall meet the requirements of MoveConstructible (Table 20). [Note:Function need not meet the requirements of CopyConstructible (Table 21).—end note]
With std::move(f)
you can be guaranteed of being able to read the mutated state externally.
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