无法推导模板参数 [英] Template parameter cannot be deduced

问题描述

我不明白为什么在这种情况下不能推断出T：

I don't understand why T cannot be deduced in this scenario:

template<class T>
class MyType
{
    T * data;
};

class MyOtherType
{
};

template<typename T>
struct MyType_OutArg
{
    typedef MyType<T> & type;
};

template<typename T>
void
DoSomething(typename MyType_OutArg<T>::type obj)
{

}

void func(MyType_OutArg<MyOtherType>::type obj)
{
    DoSomething(obj);
}

来自GCC 4.7.1，带有-std = c ++ 14

From GCC 4.7.1 with -std=c++14

<source>: In function 'void func(MyType_OutArg<MyOtherType>::type)':
26 : <source>:26:20: error: no matching function for call to 'DoSomething(MyType<MyOtherType>&)'
     DoSomething(obj);
                    ^
26 : <source>:26:20: note: candidate is:
19 : <source>:19:1: note: template<class T> void DoSomething(typename MyType_OutArg<T>::type)
 DoSomething(typename MyType_OutArg<T>::type obj)
 ^
19 : <source>:19:1: note:   template argument deduction/substitution failed:
26 : <source>:26:20: note:   couldn't deduce template parameter 'T'
     DoSomething(obj);
                    ^
Compiler returned: 1

当然，以下工作：

DoSomething<MyOtherType>(obj);

但我不确定为什么有必要。

but i'm unsure why it's necessary. Shouldn't the compiler have enough information?

推荐答案

这是因为您的情况是非推论上下文 em>。

This is because your case is a Non-deduced contexts.

引用自 http ：//en.cppreference.com/w/cpp/language/template_argument_deduction ：

非推论上下文

Non-deduced contexts

在以下情况下，用于构成P的类型，模板和非类型值不参与模板自变量的推导，而是使用在其他地方推导或明确指定。如果仅在非推导上下文中使用模板参数并且未明确指定模板参数，则模板参数推导将失败。

In the following cases, the types, templates, and non-type values that are used to compose P do not participate in template argument deduction, but instead use the template arguments that were either deduced elsewhere or explicitly specified. If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.

1）嵌套名称说明符（使用限定ID指定的类型的作用域解析运算符：:)的左侧

1) The nested-name-specifier (everything to the left of the scope resolution operator ::) of a type that was specified using a qualified-id

在您的情况下， typename MyType_OutArg< T> :: type 将不参与类型推导，并且在其他地方也不知道 T ，因此此模板功能将被忽略。

In your case, typename MyType_OutArg<T>::type will not participate in type deduction, and T is not known from elsewhere, thus this template function is ignored.

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