首页 > 解决方案 > 为什么 std::copy() 需要 std::back_inserter 插入到具有足够容量的向量中?

问题描述

参考代码:

#include <vector>
#include <algorithm>
#include <string>
#include <iostream>

void print(std::string label, std::vector<int> & arr) {
  std::cout << label << ":" << " size: " << arr.size() << " cap: " << arr.capacity() << " [ ";
  for (auto elem : arr) {
    std::cout << elem << " ";
  }
  std::cout << " ] " << std::endl;
}

void reserve_dest_use_begin() {
  std::vector<int> s_arr = {0, 1, 2, 3, 4, 5}; 
  print("source", s_arr);

  std::vector<int> d_arr;
  d_arr.reserve(3);
  print("dest", d_arr);

  auto min_elems = std::min(s_arr.size(), d_arr.capacity());

  std::cout << "COPYING FIRST" << min_elems << "3 FROM SOURCE TO DEST" << std::endl;

  std::copy(s_arr.begin(), s_arr.begin() + min_elems, d_arr.begin());

  print("source", s_arr);
  print("dest", d_arr);
}

void reserve_dest_use_back_inserter() {
  std::vector<int> s_arr = {0, 1, 2, 3, 4, 5}; 
  print("source", s_arr);

  std::vector<int> d_arr;
  d_arr.reserve(3);
  print("dest", d_arr);

  auto min_elems = std::min(s_arr.size(), d_arr.capacity());

  std::cout << "COPYING FIRST" << min_elems << " ELEMENTS FROM SOURCE TO DEST" << std::endl;

  std::copy(s_arr.begin(), s_arr.begin() + min_elems, std::back_inserter(d_arr));

  print("source", s_arr);
  print("dest", d_arr);
}

int main() {
  std::cout << "RESERVE DEST ARR. USE BEGIN() TO COPY" << std::endl;
  reserve_dest_use_begin();
  std::cout << "RESERVE DEST ARR. USE BACK_INSERTER() TO COPY" << std::endl;
  reserve_dest_use_back_inserter();

输出:

RESERVE DEST ARR USE BEGIN() TO COPY
source: size: 6 cap: 6 [ 0 1 2 3 4 5  ] 
dest: size: 0 cap: 3 [  ] 
COPYING FIRST 3 ELEMENTS FROM SOURCE TO DEST
source: size: 6 cap: 6 [ 0 1 2 3 4 5  ] 
dest: size: 0 cap: 3 [  ] 
=============================================
RESERVE DEST ARR USE BACK_INSERTER() TO COPY
source: size: 6 cap: 6 [ 0 1 2 3 4 5  ] 
dest: size: 0 cap: 3 [  ] 
COPYING FIRST 3 ELEMENTS FROM SOURCE TO DEST
source: size: 6 cap: 6 [ 0 1 2 3 4 5  ] 
dest: size: 3 cap: 3 [ 0 1 2  ]

在这两种情况下,目标阵列都有足够的容量。cppreference的文档表明:

Copies the elements in the range, defined by [first, last), to another range beginning at d_first.
1) Copies all elements in the range [first, last) starting from first and proceeding to last - 1. The behavior is undefined if d_first is within the range [first, last). In this case, std::copy_backward may be used instead.

指向的d_arr.begin()源范围之外的范围,但在提供的示例中,尽管基础向量具有足够的容量[first, last),但我需要使用std::back_inserter()复制而不是仅提供。d_arr.begin()

std::back_inserter()操作是优化为仅 memmove 内存块,还是推回每个元素?cppreference 的注释指出:

In practice, implementations of std::copy avoid multiple assignments and use bulk copy functions such as std::memmove if the value type is TriviallyCopyable and the iterator types satisfy LegacyContiguousIterator.

但是,std::back_inserter()我怀疑它不会与memmove.

总而言之,我有以下问题:

  1. 当底层向量有足够的容量时,为什么我不能d_arr.begin()用作OutputItin ?std::copy
  2. 使用是否std::back_inserter()针对批量复制范围进行了优化?

编辑: 我想我是从错误的角度提出这个问题的。评论中已经澄清了我想要做的操作是insert()而不是copy(). 我的具体用例是我反复clear()d_arr一个子向量s_arrd_arr. 我试图避免重新分配我的d_arr. 然而,因为d_arr被清除,虽然它确实有足够的容量,但它没有大小,这意味着没有要复制的元素。相反,我真正想做的是从s_arrinto插入一个子向量d_arr

d_arr.insert(d_arr.begin(), s_arr.begin(), s_arr.begin() + min_elems)

标签: c++c++11vector

解决方案


当基础向量有足够的容量时,为什么我不能在 std::copy 中使用 d_arr.begin() 作为 OutputIt?

因为目标向量是空的,因此会std::copy溢出向量(因为任何元素的赋值都超出了空向量的范围)。

std::back_inserter() 操作是否优化为仅 memmove

它可能是。它甚至可能被优化为更快的东西。

使用 std::back_inserter() 是否针对批量复制范围进行了优化?

给定足够聪明的优化器,是的。


使用适当的 vector 构造函数会比std::copy. 既适用于代码阅读者,也适用于优化者。


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