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STL algorithm算法copy(6)

 
阅读更多
原文地址:http://www.cplusplus.com/reference/algorithm/copy/
function template
<algorithm>

std::copy

template <class InputIterator, class OutputIterator>
  OutputIterator copy (InputIterator first, InputIterator last, OutputIterator result);
Copy range of elements
Copies the elements in the range[first,last)into the range beginning atresult.

复制范围[first,last)里面的元素到result的位置。(将会覆盖原来的数据)

例子:

#include <iostream>
#include <algorithm>
#include <vector>
#include <array>
using namespace std;
void copy1(){
    vector<int> v1{1,5,7,8,9};
    vector<int> v2{100,200,300};
    array<int,7> ai{888,666.555,222,111,555,777};
    vector<double> vd{9.9,8.8,7.7,};

    cout<<"at first,v1=";
    for(int &i:v1)
        cout<<i<<" ";
    cout<<endl;

    cout<<"at first,v2=";
    for(int &i:v2)
        cout<<i<<" ";
    cout<<endl;

    auto it=copy(v2.begin(),v2.end(),v1.begin());
    cout<<"after copy(v2.begin(),v2.end(),v1.begin())"<<endl;
    cout<<"v1=";
    for(int &i:v1)
        cout<<i<<" ";
    cout<<endl;

     cout<<"v2=";
    for(int &i:v2)
        cout<<i<<" ";
    cout<<endl;
    cout<<"the return it is :*it="<<*it<<endl;



}

运行截图:




The function returns an iterator to the end of the destination range (which points to the element following the last element copied).

该函数返回一个迭代器,指向目标范围的最后(即last的下一个元素)(上面例子的8)


The ranges shall not overlap in such a way thatresultpoints to an element in the range[first,last). For such cases, seecopy_backward.

result不应该指向[first,last)中的任一元素,如果需要,应该使用copy_backward.(这句话很有问题,因为copy_backward也有一句几乎一样的话,具体需求请具体分析

例子:

#include <iostream>
#include <algorithm>
#include <vector>
#include <array>
using namespace std;
void copy2(){
    vector<int> v1{1,5,7,8,9};

    cout<<"at first,v1=";
    for(int &i:v1)
        cout<<i<<" ";
    cout<<endl;


    auto it=copy(v1.begin(),v1.end(),v1.begin()+3);
    cout<<"after copy(v1.begin(),v1.end(),v1.begin()+3)"<<endl;
    cout<<"v1=";
    for(int &i:v1)
        cout<<i<<" ";
    cout<<endl;
    cout<<"the return it is :*it="<<*it<<endl;



}
运行截图:




The behavior of this function template is equivalent to:
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template<class InputIterator, class OutputIterator>
  OutputIterator copy (InputIterator first, InputIterator last, OutputIterator result)
{
  while (first!=last) {
    *result = *first;
    ++result; ++first;
  }
  return result;
}



Parameters

first, last
Input iteratorsto the initial and final positions in a sequence to be copied. The range used is[first,last), which contains all the elements betweenfirstandlast, including the element pointed byfirstbut not the element pointed bylast.
要复制的范围。
result
Output iteratorto the initial position in the destination sequence.
This shall not point to any element in the range[first,last).
要覆盖数据的开头。
不应该是[first,last)里的任一元素。

Return value

An iterator to the end of the destination range where elements have been copied.

返回一个指向目标范围最后一个元素的迭代器。


Example

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// copy algorithm example
#include <iostream>     // std::cout
#include <algorithm>    // std::copy
#include <vector>       // std::vector

int main () {
  int myints[]={10,20,30,40,50,60,70};
  std::vector<int> myvector (7);

  std::copy ( myints, myints+7, myvector.begin() );

  std::cout << "myvector contains:";
  for (std::vector<int>::iterator it = myvector.begin(); it!=myvector.end(); ++it)
    std::cout << ' ' << *it;

  std::cout << '\n';

  return 0;
}


Output:
myvector contains: 10 20 30 40 50 60 70

Complexity

Linear in thedistancebetweenfirstandlast: Performs an assignment operation for each element in the range.

Data races

The objects in the range[first,last)are accessed (each object is accessed exactly once).
The objects in the range betweenresultand the returned value are modified (each object is modified exactly once).

result以及返回值之间的所有元素都将会被修改。


Exceptions

Throws if either an element assignment or an operation on iterators throws.

Note that invalid arguments causeundefined behavior.


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//写的错误或者不好的地方请多多指导,可以在下面留言或者点击左上方邮件地址给我发邮件,指出我的错误以及不足,以便我修改,更好的分享给大家,谢谢。

转载请注明出处:http://blog.csdn.net/qq844352155

author:天下无双

Email:coderguang@gmail.com

2014-9-8

于GDUT

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