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9247114
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0
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2012-02-12T07:09:45.437
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2012-02-12T07:15:24.420
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2012-02-12T07:15:24.420
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375343
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375343
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2
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6
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text
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Generally, when lambda is used for "downward" closure, you can convert a c++ lambda to a clang block by converting: <pre><code>[&](int i) { return i; } </code></pre> to: <pre><code>^(int i) { return i; } </code></pre> There is still some subtle differences. Clang's blocks only capture C++ classes by const. I don't know if this includes C++ POD types too. Finally, if "upward" closure is needed, then the two diverge drastically. Clang's blocks require variables captured to be annotated with <code>__block</code>, which the compiler will allocate it on the heap. Whereas, in C++, the way the lambda captures, needs to be decided based on the lifetime of the object.(That is by value making a copy or by reference). Also in C++, copying the closure is handle automatically by the copy-constructor mechanism in C++. However, with clang's block, <code>Block_copy</code> and <code>Block_release</code> need to be called to handle the copying of the block. A simple wrapper can be written in C++ to handle this. For example: <pre><code>typedef void (^simple_block)(void); class block_wrapper { simple_block block; public: block_wrapper (const simple_block& x) : block(Block_copy(x)) {} void operator() () const { block(); } block_wrapper(const block_wrapper& rhs) : block(Block_copy(rhs.block)) {} block_wrapper& operator=(const block_wrapper& rhs) { if (this != &rhs) { Block_release(this->block); this->block = Block_copy(rhs.block); } return *this; } ~block_wrapper () { Block_release(this->block); } }; </code></pre>
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1. POIs it possible to convert a C++0x lambda to a clang block?
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1. USPaul Fultz II
UserOwnerUserId
1. USPaul Fultz II
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