基于其返回类型的函数模板推导?
时间：2023-04-23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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