c++ - 为什么这个模板推理失败

Question

此代码无法使用 clang++ 6.0 或 g++4.9.1 进行编译(代码没有任何意义，但这是实现它的最小示例):

#include <forward_list>

template<typename T>
T getItem(typename std::forward_list<T>::const_iterator it) {
    return *it;
}

template<typename T>
void foo() {
    std::forward_list<T> list;
    auto item = getItem(list.cbegin());
}

template<typename T>
void bar(const std::forward_list<T>& list) {
    auto item = getItem(list.cbegin());
}

int main() {
    std::forward_list<int> list;
    bar(list);
}

我收到这个错误

t2.cpp:17:17: error: no matching function for call to 'getItem'
    auto item = getItem(list.cbegin());
                ^~~~~~~
t2.cpp:22:5: note: in instantiation of function template specialization 'bar<int>' requested here
    bar(list);
    ^
t2.cpp:4:3: note: candidate template ignored: couldn't infer template argument 'T'
T getItem(typename std::forward_list<T>::const_iterator it) {
  ^
1 error generated.

要修复它，我需要像这样更改 bar() 的调用:

template<typename T>
void bar(const std::forward_list<T>& list) {
    auto item = getItem<T>(list.cbegin());
}

我不明白为什么编译器无法推断模板参数，奇怪的是编译器对 foo() 非常满意。

Answer 1

您正在尝试从非推导上下文推导模板参数，§ [temp.deduct.type]/5

The non-deduced contexts are:

— The nested-name-specifier of a type that was specified using a qualified-id.

即

template<typename T>
T getItem(typename std::forward_list<T>::const_iterator it)
                   ^^^^^^^^^^^^^^^^^^^^

和§ [temp.deduct.type]/4

In certain contexts, however, the value does not participate in type deduction, but instead uses the values of 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.

如果您尝试实例化foo it will give the same error .使用上面的代码，您不会收到 foo 的错误，因为从属名称仅在实例化时查找(这通常称为两阶段查找)。 CFR。 Semantic correctness of non-instantiated C++ template functions