类和函数在C ++中的模板类型扣除？ [英] Template type deduction in C++ for Class vs Function?

问题描述

为什么自动类型扣除只对函数而不对类才可能？

Why is that automatic type deduction is possible only for functions and not for Classes?

推荐答案

做 std :: make_pair ：

template<class T>
make_foo(T val) {
    return foo<T>(val);
}

EDIT： C ++编程语言，第三版，第335页。Bjarne说：

I just found the following in "The C++ Programming Language, Third Edition", page 335. Bjarne says:

注意类模板参数是
从来没有推导过。原因是由几个
构造函数为一个类提供的
灵活性将使
在许多
的情况下不可能进行这样的扣除，并且在更多的情况下模糊。

Note that class template arguments are never deduced. The reason is that the flexibility provided by several constructors for a class would make such deduction impossible in many cases and obscure in many more.

这当然是非常主观的。在 comp.std.c ++ 中有一些讨论，并且一致似乎是没有理由不能支持 。是否一个好主意是否是另一个问题...

This is of course very subjective. There's been some discussion about this in comp.std.c++ and the consensus seems to be that there's no reason why it couldn't be supported. Whether it would be a good idea or not is another question...

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