C++ std::next用法及代码示例

std::next在以某些no表示高级后返回指向元素的迭代器。职位。它在头文件中定义。

它不修改其自变量，并返回由指定数量增加的自变量的副本。如果它是random-access迭代器，则该函数仅使用一次运算符+或运算符-进行前进。否则，该函数在复制的迭代器上重复使用递增或递减运算符(运算符++或运算符-)，直到已推进n个元素。

用法:

ForwardIterator next (ForwardIterator it,
       typename iterator_traits::difference_type n = 1);
it:Iterator to the base position.
difference_type: It is the numerical type that represents 
distances between iterators of the ForwardIterator type.
n:Total no. of positions by which the
iterator has to be advanced. In the syntax, n is assigned
a default value 1 so it will atleast advance by 1 position.

返回：It returns an iterator to the element 
n positions away from it.

// C++ program to demonstrate std::next 
#include <iostream> 
#include <iterator> 
#include <deque> 
#include <algorithm> 
using namespace std; 
int main() 
{ 
    // Declaring first container 
    deque<int> v1 = { 1, 2, 3, 4, 5, 6, 7 }; 
  
    // Declaring another container 
    deque<int> v2 = { 8, 9, 10 }; 
  
    // Declaring an iterator 
    deque<int>::iterator i1; 
  
    // i1 points to 1 
    i1 = v1.begin(); 
  
    // Declaring another iterator to store return 
    // value and using std::next 
    deque<int>::iterator i2; 
    i2 = std::next(i1, 4); 
  
    // Using std::copy 
    std::copy(i1, i2, std::back_inserter(v2)); 
    // Remember, i1 stills points to 1 
    // and i2 points to 5 
    // v2 now contains 8 9 10 1 2 3 4 
  
    // Displaying v1 and v2 
    cout << "v1 = "; 
  
    int i; 
    for (i = 0; i < 7; ++i) { 
        cout << v1[i] << " "; 
    } 
  
    cout << "\nv2 = "; 
    for (i = 0; i < 7; ++i) { 
        cout << v2[i] << " "; 
    } 
  
    return 0; 
}

输出：

v1 = 1 2 3 4 5 6 7
v2 = 8 9 10 1 2 3 4

How can it be helpful ?

• 列表中的前进迭代器：由于list支持双向迭代器，因此只能使用++和--运算符进行递增。因此，如果我们想将迭代器前进一个以上的位置，则使用std::next可能非常有用。

  // C++ program to demonstrate std::next 
  #include <iostream> 
  #include <iterator> 
  #include <list> 
  #include <algorithm> 
  using namespace std; 
  int main() 
  { 
      // Declaring first container 
      list<int> v1 = { 1, 2, 3, 7, 8, 9 }; 
    
      // Declaring second container 
      list<int> v2 = { 4, 5, 6 }; 
    
      list<int>::iterator i1; 
      i1 = v1.begin(); 
      // i1 points to 1 in v1 
    
      list<int>::iterator i2; 
      // i2 = v1.begin() + 3; 
      // This cannot be used with lists 
      // so use std::next for this 
    
      i2 = std::next(i1, 3); 
    
      // Using std::copy 
      std::copy(i1, i2, std::back_inserter(v2)); 
      // v2 now contains 4 5 6 1 2 3 
    
      // Displaying v1 and v2 
      cout << "v1 = "; 
    
      int i; 
      for (i1 = v1.begin(); i1 != v1.end(); ++i1) { 
          cout << *i1 << " "; 
      } 
    
      cout << "\nv2 = "; 
      for (i1 = v2.begin(); i1 != v2.end(); ++i1) { 
          cout << *i1 << " "; 
      } 
    
      return 0; 
  }

  输出：

  v1 = 1 2 3 7 8 9
  v2 = 4 5 6 1 2 3  
  

  说明：在这里，请看一下如何只复制列表的选定部分，然后才能使用std::next，否则我们不能使用带有列表支持的双向迭代器的+ =，-=运算符。因此，我们使用了std::next并将迭代器直接前进了三个位置。