llvm :: Module是两个运算符。*和-> *的左侧

Question

I was am reading LLVM Tutorial and saw these statements (at different positions):

static std::unique_ptr<Module> TheModule;
TheModule.get();
TheModule->getFunction(arg);

But when I try putting this in my code:

TheModule->get();

I get error:

myTest.cpp:116:16: error: no member named 'get' in 'llvm::Module' TheModule->get();

Why could llvm::Module be the left side of both .* and ->* ? And why does TheModule.get() work, but TheModule->get() doesn't?

Does this have anything to do with std::unique_ptr ?

Answer 1

Does this have anything to do with std::unique_ptr?

Yes, it does.

It's the std::unique_ptr 's get you managed to call with TheModule.get() , which returns the managed pointer, TheModule->get() is not the same, it's equivalent to:

TheModule.get()->get();
(*TheModule).get();
(*TheModule.get()).get();

These are all the same and they call get on the managed llvm::Module instance. Of course, llvm::Module doesn't have get , that's why you're getting the error.

It's really not that surprising once you learn what operator* and operator-> of std::unique_ptr do. These operators exist, so you can work with smart pointers (syntactically) the same way as with raw ones. And std::unique_ptr::get returns a raw pointer to the managed object.

Answer 2

When you use the "arrow" operator -> you access the contained pointer, when you use the dot operator . you access the unique_ptr object.

For example

TheModule.get()  // Calls std::unique_ptr<T>::get()

TheModule->getFunction(arg);  // Calls llvm::Module::getFunction()