“基本C++” ch3.9使用 “back_inserter（ivec2）” 取代 “i​​vec2.begin（）” 不能编译

Question

这个例子从书中复制：

#include "stdafx.h" 
#include<functional> 
#include<iostream> 
#include <vector> 
#include <algorithm> 
#include<iterator> 

using namespace std; 

template<typename input_iterator_tag, typename output_iterator_tag, 
typename elem_type, typename comp> 
output_iterator_tag 
filter(input_iterator_tag first, output_iterator_tag last, 
output_iterator_tag at, const elem_type &val, comp pred) 
{ 
    while ((first = find_if(first, last, bind2nd(pred, val))) != last) 
    { 
     cout << "found value: " << *first << endl; 
     *at++ = *first++; 
    } 
    return at; 
} 

int _tmain(int argc, _TCHAR* argv[]) 
{ 
const int elem_size = 8; 
int ia[elem_size] = { 12, 8, 43, 0, 6, 21, 3, 7 }; 
vector<int>ivec(ia, ia + elem_size); 

int ia2[elem_size]; 
vector<int>ivec2(elem_size); 

cout << "filtering integer array for values less than 8\n"; 
filter(ia, ia + elem_size, ia2, elem_size, less<int>()); 
cout << "filtering integer vector for value greater than 8\n"; 
filter(ivec.begin(), ivec.end(), ivec2.begin(), elem_size, greater<int> ());//it's work ok 
filter(ivec.begin(), ivec.end(), back_inserter(ivec2), elem_size, greater<int>());//compile error,no instance of function template "filter" matches 

system("pause"); 
return 0; 
} 

我为什么用“back_inserter（ ivec2）“取代”ivec2.begin（）“，但它不能在visual studio中编译;

IntelliSense: no instance of function template "filter" matches the argument list argument types are: (std::_Vector_iterator<std::_Vector_val<std::_Simple_types<int>>>, std::_Vector_iterator<std::_Vector_val<std::_Simple_types<int>>>, std::back_insert_iterator<std::vector<int, std::allocator<int>>>, const int, std::greater<int>) 

任何人都可以帮我纠正它，为什么？

Answer 1

看起来像函数声明中的拼写错误。第二个参数（last）应该是input_iterator_tag而不是output_iterator_tag，因为它应该与first迭代器相匹配，因为它将直接与first进行比较。

template<typename input_iterator_tag, typename output_iterator_tag, 
typename elem_type, typename comp> 
output_iterator_tag 
filter(input_iterator_tag first, input_iterator_tag last, 
output_iterator_tag at, const elem_type &val, comp pred) 

注意：它为您的第一个案例工作的原因是，输入和输出类型都是相同的。