无法推导模板类型 [英] Can't deduce template type

问题描述

我试图传递一个迭代器作为模板参数到模板方法，但编译器抱怨：

I'm trying to pass an iterator as a template parameter to a template method, but the compiler complains that:

error C2783: 'void Test::Assert(std::vector<T>::const_iterator)':
could not deduce template argument for 'T'

产生错误的代码是：

#include "stdafx.h"
#include <iostream>
#include <vector>

class Test
{
    public:
        template <typename T>
        void Assert(typename std::vector<T>::const_iterator it)
        {
            std::cout << *it << std::endl;
        }
};

int _tmain(int argc, _TCHAR* argv[])
{
    Test test;

    std::vector<double> myVec;

    test.Assert(myVec.cbegin());

    return 0;
}

我猜想有一个简单的方法使这项工作，因为大多数的std算法可以从迭代器中推导出类型。

I'm guessing there is a simple way to make this work, since most of the std algorithms can deduce type from iterator.

推荐答案

原因是你有 T in是未推导的上下文：

template <typename T>
void Assert(typename std::vector<T>::const_iterator it)

一个更简单的例子来理解为什么：

Consider a simpler case to understand why:

struct A { using type = int; };
struct B { using type = int; };
struct C { using type = int; };

template <typename T>
void Assert(typename T::type it) { ... }

Assert(5);

T 这是不可能确定。您必须显式提供类型为 Assert< A>（5）。

What should T deduce as? It's impossible to determine. You'd have to explicitly provide the type... as something like Assert<A>(5).

另请参见什么是不受限制的上下文？

因为大多数std算法可以从迭代器中推导出类型。

since most of the std algorithms can deduce type from iterator.

$ b b

这是因为标准算法只是推导出迭代器类型，而不是容器类型。例如 std :: find is just：

That's because the standard algorithms just deduce the iterator type, not the container type. For instance std::find is just:

template <class InputIt, class T>
InputIt find( InputIt first, InputIt last, const T& value );

这里没有container的概念 - 它只是迭代器类型，需要推断。这是算法库的美的一部分。

There is no concept of "container" here at all - it's just the iterator type that needs to be deduced. That's part of the beauty of the algorithms library.

所以如果你想做的只是输出迭代器的内容，正确的函数只是：

So if what you want to do is just output the contents of the iterator, the correct function would just be:

template <typename Iterator>
void Assert(Iterator it)
{
    std::cout << *it << std::endl;
}

当调用 Assert（myVec.cbegin ）， Iterator 将会被推导为 std :: vector< double> :: const_iterator ，这正是你想要的。

When you call Assert(myVec.cbegin()), Iterator will get deduced as std::vector<double>::const_iterator, which is exactly what you want.

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