使用模板时的类型推断 [英] type inference when using templates
问题描述
所以这里是我想做的:我使用 std :: pair
,但我一定会喜欢做同样使用元组,或者几乎任何种类的模板。当分配一个对变量时,我需要输入如下:
So here is what I would like to do: I use std::pair
, but I would surely like to do the same using tuples, or indeed pretty much any kind of template. When assigning a pair variable, I need to type something like:
T1 t1;
T2 t2;
std::pair<T1,T2> X;
X = std::pair<T1,T2> (t1, t2);
有没有办法省略第二个< T1,T2& / code>创建一个新对,并让编译器猜测,使用X的类型(我显然是试图创建一个
对< T1,T2>
)或 t1
和 t2
的类型(我正在构建一个 / code>对象和
T2
对象,有一个机会,我想要的对类型 pair< T1,T2> code>)?
Is there a way to omit the second <T1,T2>
when creating the new pair, and let the compiler guess, either using X's type (I'm obviously trying to create a pair<T1,T2>
) or t1
and t2
's types (I am building a pair with a T1
object and a T2
object, there is a chance the pair I want is of type pair<T1,T2>
) ?
推荐答案
是的,但模板参数扣除只适用于函数模板,而不适用于类模板的构造函数。因此,库提供了一个函数:
Yes, but template argument deduction only works on function templates, not constructors of class templates. For this reason, the library provides a function:
X = std::make_pair(t1, t2);
在C ++ 11中,对和元组可以初始化并从初始化器列表中分配:
In C++11, pairs and tuples can be initialised and assigned from initialiser lists:
X = {t1, t2};
或 auto
类型从初始化程序,所以你不需要指定模板参数:
or auto
can be used to automatically specify the type from the initialiser, so you don't need to specify the template arguments at all:
auto X = std::make_pair(t1, t2);
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