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/**
* Licensed to the Apache Software Foundation (ASF) under one
* or more contributor license agreements. See the NOTICE file
* distributed with this work for additional information
* regarding copyright ownership. The ASF licenses this file
* to you under the Apache License, Version 2.0 (the
* "License"); you may not use this file except in compliance
* with the License. You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing,
* software distributed under the License is distributed on an
* "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY
* KIND, either express or implied. See the License for the
* specific language governing permissions and limitations
* under the License.
*/
#ifndef ScopedLambda_h
#define ScopedLambda_h
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<MyThings* (int, Stuff&)>&);
//
// The caller just does:
//
// void foo(scopedLambda<MyThings* (int, Stuff&)>([&] (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<typename FunctionType> class ScopedLambda;
template<typename ResultType, typename... ArgumentTypes>
class ScopedLambda<ResultType (ArgumentTypes...)> {
public:
ScopedLambda(ResultType (*impl)(void* arg, ArgumentTypes...) = nullptr, void* arg = nullptr)
: m_impl(impl)
, m_arg(arg)
{
}
template<typename... PassedArgumentTypes>
ResultType operator()(PassedArgumentTypes&&... arguments) const
{
return m_impl(m_arg, std::forward<PassedArgumentTypes>(arguments)...);
}
private:
ResultType (*m_impl)(void* arg, ArgumentTypes...);
void *m_arg;
};
template<typename FunctionType, typename Functor> class ScopedLambdaFunctor;
template<typename ResultType, typename... ArgumentTypes, typename Functor>
class ScopedLambdaFunctor<ResultType (ArgumentTypes...), Functor> : public ScopedLambda<ResultType (ArgumentTypes...)> {
public:
template<typename PassedFunctor>
ScopedLambdaFunctor(PassedFunctor&& functor)
: ScopedLambda<ResultType (ArgumentTypes...)>(implFunction, this)
, m_functor(std::forward<PassedFunctor>(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<ResultType (ArgumentTypes...)>(implFunction, this)
, m_functor(other.m_functor)
{
}
ScopedLambdaFunctor(ScopedLambdaFunctor&& other)
: ScopedLambda<ResultType (ArgumentTypes...)>(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<ScopedLambdaFunctor*>(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<Foo(Bar)>([&] (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<typename FunctionType, typename Functor>
ScopedLambdaFunctor<FunctionType, Functor> scopedLambda(const Functor& functor)
{
return ScopedLambdaFunctor<FunctionType, Functor>(functor);
}
template<typename FunctionType, typename Functor>
ScopedLambdaFunctor<FunctionType, Functor> scopedLambda(Functor&& functor)
{
return ScopedLambdaFunctor<FunctionType, Functor>(WTFMove(functor));
}
template<typename FunctionType, typename Functor> class ScopedLambdaRefFunctor;
template<typename ResultType, typename... ArgumentTypes, typename Functor>
class ScopedLambdaRefFunctor<ResultType (ArgumentTypes...), Functor> : public ScopedLambda<ResultType (ArgumentTypes...)> {
public:
ScopedLambdaRefFunctor(const Functor& functor)
: ScopedLambda<ResultType (ArgumentTypes...)>(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<ResultType (ArgumentTypes...)>(implFunction, this)
, m_functor(other.m_functor)
{
}
ScopedLambdaRefFunctor(ScopedLambdaRefFunctor&& other)
: ScopedLambda<ResultType (ArgumentTypes...)>(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<ScopedLambdaRefFunctor*>(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.
template<typename FunctionType, typename Functor>
ScopedLambdaRefFunctor<FunctionType, Functor> scopedLambdaRef(const Functor& functor)
{
return ScopedLambdaRefFunctor<FunctionType, Functor>(functor);
}
} // namespace WTF
using WTF::ScopedLambda;
using WTF::scopedLambda;
using WTF::scopedLambdaRef;
#endif // ScopedLambda_h