Question

我正在尝试实现一个简单的模板函数，这段代码不能编译，但我希望它会让你知道我正在尝试做什么：

template<typename T>
typename T::iterator   do_find(const T& container, const int val)
{
   return std::find(container.begin(), container.end(), val);
}

我想返回发现自己返回的迭代器，而不知道我在模板函数do_find中收到的容器类型。我做错了什么？

这是主要的：

int             main()
{
  std::vector<int>      c;

  c.push_back(42);
  c.push_back(0);
  c.push_back(1);
  c.push_back(-58);
  c.push_back(42);
  c.push_back(777);
  c.push_back(1911);
  c.push_back(9);

  do_find(c, 42);

  return 0;
}

编译错误：

In file included from main.cpp:14:0:
find.hpp: In instantiation of ‘typename T::iterator do_find(const T&, int) [with T = std::vector<int>; typename T::iterator = __gnu_cxx::__normal_iterator<int*, std::vector<int> >]’:
main.cpp:29:16:   required from here
find.hpp:17:59: error: could not convert ‘std::find<__gnu_cxx::__normal_iterator<const int*, std::vector<int> >, int>((& container)->std::vector<_Tp, _Alloc>::begin<int, std::allocator<int> >(), (& container)->std::vector<_Tp, _Alloc>::end<int, std::allocator<int> >(), (* & val))’ from ‘__gnu_cxx::__normal_iterator<const int*, std::vector<int> >’ to ‘std::vector<int>::iterator {aka __gnu_cxx::__normal_iterator<int*, std::vector<int> >}’
find.hpp: In function ‘typename T::iterator do_find(const T&, int) [with T = std::vector<int>; typename T::iterator = __gnu_cxx::__normal_iterator<int*, std::vector<int> >]’:
find.hpp:18:1: error: control reaches end of non-void function [-Werror=return-type]
cc1plus: all warnings being treated as errors

Answer 1

如果您有const T& container作为此功能的第一个参数，则必须返回typename T::const_iterator。

在模板中返回一个迭代器

1 个答案: