C++ 模板不接受迭代器

2022-01-10 00:00:00 quicksort iterator templates c++

我正在重新学习 C++，并开始尝试应该是一个简单的算法:快速排序.我的函数有这个签名:

I'm re-learning C++, and have started by trying what should be a simple algorithm: QuickSort. My function has this signature:

template <class T>
void QSort(typename std::vector<T>::iterator begin, typename std::vector<T>::iterator end)

它在我的主函数中被调用:

And it is called in my main function:

int main()
{
    std::vector<int> unsort({56,32,11,45,67,81,12,5});
    std::vector<int>::iterator b=unsort.begin();
    std::vector<int>::iterator e=unsort.end();
    QSort(b, e);
    return 0;
}

并给出此错误:

C:UsersDeusProjectsQSortmain.cpp||In function 'int main()':|
C:UsersDeusProjectsQSortmain.cpp|49|error: no matching function for call to 'QSort(__gnu_cxx::__normal_iterator<int*, std::vector<int, std::allocator<int> > >&, __gnu_cxx::__normal_iterator<int*, std::vector<int, std::allocator<int> > >&)'|
||=== Build finished: 1 errors, 0 warnings ===|

编译器似乎无法解析 T 应该是什么.有没有办法做我想做的事情，或者我应该将参数声明为 T 类型，并处理由此产生的不确定性?

It seems that the compiler is having trouble resolving what T should be. Is there a way to do what I'm trying to do, or should I just declare the arguments as type T, and work with the resulting uncertainty?

推荐答案

编译器无法从你的函数调用中推断出 T.想想当 std::vector<T>::iterator 是 T*:

The compiler has no way to deduce T from your function call. Think about what happens when std::vector<T>::iterator is T*:

int *b = ...;
int *e = ...;
QSort(b, e);

一般来说，如果你写typenameSomething::anotherThing，那么TemplateParemter在调用中是无法推导出来的.必须明确提供

In general, if you write typename Something<TemplateParameter>::anotherThing, then the TemplateParemter cannot be deduced in the call. It must be explicitly provided

QSort<int>(b, e);

我建议只使用 T 作为参数类型.这将使您不仅可以接受向量迭代器，还可以接受 T* 或 std::deque<T>::iterator 和任何其他随机访问迭代器.

I recommend to just use T as the parameter type. That will allow you to not only accept vector iterators, but also T*, or std::deque<T>::iterator and any other random access iterators.

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