模板参数推断失败

Question

考虑以下代码：

template<class T>
vector<T> filter(typename vector<T>::iterator begin,
          typename vector<T>::iterator end,
          bool (*cond)(T a))
{
    vector<T> vec;
    for (typename vector<T>::iterator it = begin; it != end; it++) {
        if (cond(*it)) {
            vec.push_back(*it);
        }
    }
    return vec;
}

vector<int> vec = { 1,2,3,4,5,6,7,8,9,10 };
auto another_vec = filter<int>(vec.begin(), vec.end(), [](int a) {return a > 5; });

当我从函数过滤器的调用中删除类型时，代码无法编译，即写

时

filter(vec.begin(), vec.end(), [](int a) {return a > 5; });

我的问题是，为什么？编译器可以从 lambda 和迭代器中推导出类型。

我得到的错误是：

Error C2784 'std::vector> filter(vector>::iterator,vector>::iterator,bool (__cdecl *)(T))': could not deduce template argument for 'bool (__cdecl *)(T)' from 'main::' example c:\users\danii\documents\visual studio 2017\projects\example\example\source.cpp 24

我找不到有关此问题的详细信息。我的猜测是，编译器无法推断出内部类型？（例如不能从 vector 推导出 int）。如果是这样，为什么会这样？如果不是，情况如何？有什么办法可以解决吗？

我遇到的另一件事是使用迭代器本身作为模板，即类似于

template <class T, class iter, class cond>
vector<T> filter(iter begin, iter end, cond c)

编程正确吗？我觉得这段代码有点可疑。

Answer 1

My guess is, the compiler can't deduce inner types? (e.g can't deduce int from vector).

是的。这属于 non-deduced contexts:

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

请注意，没有捕获的 lambda 表达式（第 3 个参数）可以隐式转换为函数指针，但 template argument deduction 不考虑隐式转换。

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

那么这里类型推导失败

您的修复想法是个好主意，但您不需要模板参数 T，它不能（也不需要）推导出来。您可以将其更改为：

template<class iter, class cond>
auto filter(iter begin, iter end, cond c)
{
    vector<typename std::iterator_traits<iter>::value_type> vec;
    for (auto it = begin; it != end; it++) {
        if (cond(*it)) {
            vec.push_back(*it);
        }
    }
    return vec;
}

Answer 2

这是因为 lambda 类型不完全是预期的函数类型，所以模板推导将不起作用。编译器不能同时进行隐式转换和模板推导。如果你使用一个函数，它会：

bool f(int a) {
    return a > 5;
}

int main() {
    vector<int> vec = { 1,2,3,4,5,6,7,8,9,10 };
    auto another_vec = filter(vec.begin(), vec.end(), f); 
    return 0;
}

模板参数推断失败

Template parameter inference fails

c++

templates

template-argument-deduction