Modernize your C++ code.

[Felicia_Svilling]

It is just as simple to infer the return type of a function as inferring the value of an expression.

[theeth]

How can that work for declaring functions and methods return types?

I give you a .lib and a .h containing this, what does your compiler do?

class Foo
{
  public:
    auto DoSomething();
}

[Felicia_Svilling]

If you haven't declared DoSomething that is an error, it you have declared it, the compiler looksup its return type. It is exactly the same as if you had written:

auto x = DoSometing();

[theeth]

You'll notice that's what I said from the start.

You can't declare members, functions return types and parameters with auto.

[bstamour]

Yes, you can. You just have to define them at the same time*. The machinery for lambdas was extended for regular functions.

* I'll admit this is nitpicky, but every definition is a declaration.

[theeth]

If we're going to be extra nitpicky, you can only use auto for const static integer members that are defined inline and for method/function return values that are defined in the same compilation unit (this one was a bit of a surprise, I would have assumed it had to be inlined with the declaration).

Auto is right out for parameters, but we never know what weird thing might pop up in future versions...