Scala和C ++ 11的类型推断有什么区别？

Question

I'm curious what the differences between the type inference of Scala and C++11 are. In which situations do I have to specify the types in one language but not in the other? One difference seems to be the return type of functions which always have to be specified in C++11, although decltype and the new function syntax with a trailing return type allow to specify a inferred type.

Answer 1

C++ cannot infer such anonymous functions:

// this wont work*
void somefunc(std::vector<int>& v) 
{
    std::for_each(v.begin(), v.end(), [](auto &x) { x++; });
}
//                                        /\
//                                         ------ I want this to be infered

whereas Scala can:

def somefunc(v: Vector[Int]) = v.map(x => x +1)

* not sure that I've dealt correctly with C++ code syntax, I don't insult language, but it's really cryptic. If I've made a mistake, correct me, please

Answer 2

In essence, C++ inference is simplistic compared to the grown ups.

Functional languages generally something close to Hindley/Milner which is pretty close to solving an equation systems and allow to have unknowns on both sides of the fence.

On the contrary, C++ expect to be able to know the type of any inner expression and from that deduce the type of the outer expression. It's a strictly one way inference, meaning that:

auto x = foo(1, 2);

works as expected as long as there is a foo accepting integers and returning non-void. However, as demonstrated by om-nom-nom:

foo(1, [](auto x) { ++x; });

Will not work, because you cannot go backward and use the purported type of foo to deduce the type of the lambda.

The reason behind is that C++ uses overloads of functions, meaning that several definitions of foo could exist, and you actually need to know the types of the arguments to elect the right one. Since in general the above expression would be undecidable, it is forbidden even in limited cases it could have been allowed to avoid future maintenance hell and people never knowing when or not it can be used.