197 lines
7.0 KiB
C++
197 lines
7.0 KiB
C++
/*
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* Copyright (C) 2015-2016 Apple Inc. All rights reserved.
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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*
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* THIS SOFTWARE IS PROVIDED BY APPLE INC. ``AS IS'' AND ANY
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* EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR
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* PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL APPLE INC. OR
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* CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL,
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* EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO,
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* PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR
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* PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY
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* OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
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* (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
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* OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
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*/
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#pragma once
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#include <wtf/ForbidHeapAllocation.h>
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namespace WTF {
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// You can use ScopedLambda to efficiently pass lambdas without allocating memory or requiring
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// template specialization of the callee. The callee should be declared as:
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//
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// void foo(const ScopedLambda<MyThings* (int, Stuff&)>&);
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//
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// The caller just does:
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//
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// void foo(scopedLambda<MyThings* (int, Stuff&)>([&] (int x, Stuff& y) -> MyThings* { blah }));
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//
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// Note that this relies on foo() not escaping the lambda. The lambda is only valid while foo() is
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// on the stack - hence the name ScopedLambda.
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template<typename FunctionType> class ScopedLambda;
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template<typename ResultType, typename... ArgumentTypes>
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class ScopedLambda<ResultType (ArgumentTypes...)> {
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WTF_FORBID_HEAP_ALLOCATION;
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public:
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ScopedLambda(ResultType (*impl)(void* arg, ArgumentTypes...) = nullptr, void* arg = nullptr)
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: m_impl(impl)
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, m_arg(arg)
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{
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}
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template<typename... PassedArgumentTypes>
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ResultType operator()(PassedArgumentTypes&&... arguments) const
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{
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return m_impl(m_arg, std::forward<PassedArgumentTypes>(arguments)...);
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}
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private:
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ResultType (*m_impl)(void* arg, ArgumentTypes...);
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void *m_arg;
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};
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template<typename FunctionType, typename Functor> class ScopedLambdaFunctor;
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template<typename ResultType, typename... ArgumentTypes, typename Functor>
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class ScopedLambdaFunctor<ResultType (ArgumentTypes...), Functor> : public ScopedLambda<ResultType (ArgumentTypes...)> {
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public:
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template<typename PassedFunctor>
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ScopedLambdaFunctor(PassedFunctor&& functor)
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: ScopedLambda<ResultType (ArgumentTypes...)>(implFunction, this)
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, m_functor(std::forward<PassedFunctor>(functor))
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{
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}
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// We need to make sure that copying and moving ScopedLambdaFunctor results in a ScopedLambdaFunctor
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// whose ScopedLambda supertype still points to this rather than other.
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ScopedLambdaFunctor(const ScopedLambdaFunctor& other)
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: ScopedLambda<ResultType (ArgumentTypes...)>(implFunction, this)
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, m_functor(other.m_functor)
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{
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}
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ScopedLambdaFunctor(ScopedLambdaFunctor&& other)
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: ScopedLambda<ResultType (ArgumentTypes...)>(implFunction, this)
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, m_functor(WTFMove(other.m_functor))
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{
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}
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ScopedLambdaFunctor& operator=(const ScopedLambdaFunctor& other)
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{
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m_functor = other.m_functor;
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return *this;
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}
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ScopedLambdaFunctor& operator=(ScopedLambdaFunctor&& other)
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{
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m_functor = WTFMove(other.m_functor);
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return *this;
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}
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private:
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static ResultType implFunction(void* argument, ArgumentTypes... arguments)
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{
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return static_cast<ScopedLambdaFunctor*>(argument)->m_functor(arguments...);
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}
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Functor m_functor;
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};
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// Can't simply rely on perfect forwarding because then the ScopedLambdaFunctor would point to the functor
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// by const reference. This would be surprising in situations like:
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//
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// auto scopedLambda = scopedLambda<Foo(Bar)>([&] (Bar) -> Foo { ... });
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//
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// We expected scopedLambda to be valid for its entire lifetime, but if it computed the lambda by reference
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// then it would be immediately invalid.
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template<typename FunctionType, typename Functor>
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ScopedLambdaFunctor<FunctionType, Functor> scopedLambda(const Functor& functor)
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{
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return ScopedLambdaFunctor<FunctionType, Functor>(functor);
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}
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template<typename FunctionType, typename Functor>
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ScopedLambdaFunctor<FunctionType, Functor> scopedLambda(Functor&& functor)
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{
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return ScopedLambdaFunctor<FunctionType, Functor>(std::forward<Functor>(functor));
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}
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template<typename FunctionType, typename Functor> class ScopedLambdaRefFunctor;
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template<typename ResultType, typename... ArgumentTypes, typename Functor>
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class ScopedLambdaRefFunctor<ResultType (ArgumentTypes...), Functor> : public ScopedLambda<ResultType (ArgumentTypes...)> {
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public:
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ScopedLambdaRefFunctor(const Functor& functor)
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: ScopedLambda<ResultType (ArgumentTypes...)>(implFunction, this)
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, m_functor(&functor)
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{
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}
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// We need to make sure that copying and moving ScopedLambdaRefFunctor results in a
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// ScopedLambdaRefFunctor whose ScopedLambda supertype still points to this rather than
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// other.
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ScopedLambdaRefFunctor(const ScopedLambdaRefFunctor& other)
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: ScopedLambda<ResultType (ArgumentTypes...)>(implFunction, this)
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, m_functor(other.m_functor)
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{
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}
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ScopedLambdaRefFunctor(ScopedLambdaRefFunctor&& other)
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: ScopedLambda<ResultType (ArgumentTypes...)>(implFunction, this)
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, m_functor(other.m_functor)
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{
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}
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ScopedLambdaRefFunctor& operator=(const ScopedLambdaRefFunctor& other)
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{
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m_functor = other.m_functor;
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return *this;
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}
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ScopedLambdaRefFunctor& operator=(ScopedLambdaRefFunctor&& other)
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{
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m_functor = other.m_functor;
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return *this;
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}
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private:
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static ResultType implFunction(void* argument, ArgumentTypes... arguments)
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{
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return (*static_cast<ScopedLambdaRefFunctor*>(argument)->m_functor)(arguments...);
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}
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const Functor* m_functor;
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};
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// This is for when you already refer to a functor by reference, and you know its lifetime is
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// good. This just creates a ScopedLambda that points to your functor.
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//
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// Note that this is always wrong:
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//
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// auto ref = scopedLambdaRef([...] (...) {...});
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//
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// Because the scopedLambdaRef will refer to the lambda by reference, and the lambda will die after the
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// semicolon. Use scopedLambda() in that case.
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template<typename FunctionType, typename Functor>
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ScopedLambdaRefFunctor<FunctionType, Functor> scopedLambdaRef(const Functor& functor)
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{
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return ScopedLambdaRefFunctor<FunctionType, Functor>(functor);
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}
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} // namespace WTF
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using WTF::ScopedLambda;
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using WTF::scopedLambda;
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using WTF::scopedLambdaRef;
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