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