基于其返回类型的函数模板推导? [英] Template deduction for function based on its return type?

问题描述

我希望能够使用模板推导来实现以下目标:

I'd like to be able to use template deduction to achieve the following:

GCPtr<A> ptr1 = GC::Allocate();
GCPtr<B> ptr2 = GC::Allocate();

而不是(我目前拥有的):

instead of (what I currently have):

GCPtr<A> ptr1 = GC::Allocate<A>();
GCPtr<B> ptr2 = GC::Allocate<B>();

我当前的 Allocate 函数如下所示:

My current Allocate function looks like this:

class GC
{
public:
    template <typename T>
    static GCPtr<T> Allocate();
};

这能去掉额外的和吗?

推荐答案

那是不可能的.返回类型不参与类型推导，而是已经匹配了适当的模板签名的结果.不过，您可以将其隐藏在大多数用途中:

That cannot be done. The return type does not take part in type deduction, it is rather a result of having already matched the appropriate template signature. You can, nevertheless, hide it from most uses as:

// helper
template <typename T>
void Allocate( GCPtr<T>& p ) {
   p = GC::Allocate<T>();
}

int main()
{
   GCPtr<A> p = 0;
   Allocate(p);
}

该语法实际上比最初的 GCPtr 好还是差?p = GC::Allocate() 是另一个问题.

Whether that syntax is actually any better or worse than the initial GCPtr<A> p = GC::Allocate<A>() is another question.

附言c++11 将允许您跳过类型声明之一:

P.S. c++11 will allow you to skip one of the type declarations:

auto p = GC::Allocate<A>();   // p is of type GCPtr<A>

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