[英]How does this template type deduction and overload resolution work?
#include <iterator>
#include <algorithm>
#include <iostream>
#include <vector>
template <typename container>
void sort(typename container::iterator beginning,
typename container::iterator end)
{
std::cout << "calling custom sorting function\n";
}
int main()
{
std::vector<int> v{1, 2, 3};
sort(v.begin(), v.end());
}
It is gonna call the std::sort
function, which is found by ADL. 它将调用
std::sort
函数,该函数由ADL发现。 Though the code below: 虽然代码如下:
#include <iterator>
#include <algorithm>
#include <iostream>
#include <vector>
template <typename Iterator>
void sort(Iterator beginning,
Iterator end)
{
std::cout << "calling custom sorting function\n";
}
int main()
{
std::vector<int> v{1, 2, 3};
sort(v.begin(), v.end());
}
Causes ambiguous overload error. 导致模糊的过载错误。 So I have two questions:
所以我有两个问题:
How container
was deduced in code #1? 如何在代码#1中推断出
container
?
From what I know, type deduction during template instantiation cannot backtrack through member types to find the enclosing one ( std::vector<int>
in this case). 据我所知,模板实例化期间的类型推导不能回溯成员类型以找到封闭的(在这种情况下为
std::vector<int>
)。
Even if it could backtrack, why did it compile without causing ambiguous overload error? 即使它可以回溯,为什么编译时不会导致模糊的过载错误?
How
container
was deduced in code #1?如何在代码#1中推断出
container
?
It can't be deduced during template argument deduction because of non-deduced context , 由于非推断的上下文 ,在模板参数推导期间无法推导出它,
1) The nested-name-specifier (everything to the left of the scope resolution operator
::
) of a type that was specified using a qualified-id:1)使用qualified-id指定的类型的嵌套名称说明符(作用域解析运算符
::
左侧的所有内容):
That means your sort
won't be considered for overload resolution at all, then std::sort
is called without ambiguity. 这意味着您的
sort
根本不会被考虑用于重载解析,然后调用std::sort
而不会产生歧义。
Code #2 doesn't have such issue, both your sort
and std::sort
are valid candidates then leads to ambiguous overloading error. 代码#2没有这样的问题,你的
sort
和std::sort
都是有效的候选者,然后导致模糊的重载错误。
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