Question

我有以下玩具示例：

#include<iostream>
#include<vector>


int main(){


    std::vector<int> a={1, 2, 3};
    std::vector<int> b={4, 5, 6};

    int* pa = a.data();
    int* pb = b.data();

    std::swap(pa,pb);
    std::cout<<"after std::swap(pa,pb)\n";
    std::cout<<"a= "<<a[0]<<" "<<a[1]<<" "<<a[2]<<"\n";
    std::cout<<"b= "<<b[0]<<" "<<b[1]<<" "<<b[2]<<"\n";

    std::cout<<"pa= "<<pa[0]<<" "<<pa[1]<<" "<<pa[2]<<"\n";
    std::cout<<"pb= "<<pb[0]<<" "<<pb[1]<<" "<<pb[2]<<"\n";

    std::swap(a,b);

    std::cout<<"after std::swap(a,b)\n";
    std::cout<<"a= "<<a[0]<<" "<<a[1]<<" "<<a[2]<<"\n";
    std::cout<<"b= "<<b[0]<<" "<<b[1]<<" "<<b[2]<<"\n";

    std::cout<<"pa= "<<pa[0]<<" "<<pa[1]<<" "<<pa[2]<<"\n";
    std::cout<<"pb= "<<pb[0]<<" "<<pb[1]<<" "<<pb[2]<<"\n";
}

交换之后会发生什么（pa，pb）对我来说很清楚，我希望pa指向b.data（），但是在交换（a，b）之后，我希望b.data（）现在指向＆＃34 ; 1 2 3＆＃34;，但打印出pa [0]，pa [1]，pa [2] ...时不是这种情况......

Answer 1

http://en.cppreference.com/w/cpp/container/vector/swap注意到：

所有迭代器和引用仍然有效。过去的迭代器无效。

这意味着内存分配保持不变，矢量字段交换

  std::vector<int> a={1, 2, 3}; // data lays at pa
  std::vector<int> b={4, 5, 6}; // data lays at pb

  int* c = a.data(); // == pa
  int* d = b.data(); // == pb

  std::swap(c,d); // c == pb, d == pa
  std::cout<<"after std::swap(c,d)\n";
  std::cout<<"a= "<<a[0]<<" "<<a[1]<<" "<<a[2]<<"\n";
  std::cout<<"b= "<<b[0]<<" "<<b[1]<<" "<<b[2]<<"\n";

  std::cout<<"c= "<<c[0]<<" "<<c[1]<<" "<<c[2]<<"\n";
  std::cout<<"d= "<<d[0]<<" "<<d[1]<<" "<<d[2]<<"\n";

  std::swap(a,b); // c == pb, d == pa, a == pb, b == pa

  std::cout<<"after std::swap(a,b)\n";
  std::cout<<"a= "<<a[0]<<" "<<a[1]<<" "<<a[2]<<"\n";
  std::cout<<"b= "<<b[0]<<" "<<b[1]<<" "<<b[2]<<"\n";

  std::cout<<"c= "<<c[0]<<" "<<c[1]<<" "<<c[2]<<"\n";
  std::cout<<"d= "<<d[0]<<" "<<d[1]<<" "<<d[2]<<"\n";

Answer 2

向量倾向于使用指向数据的指针来实现。所以，假设内存中有一个名为foo的地方，其中是{1, 2, 3}和bar的起始位置，其中 {4, 5, 6}的起始位置

std::vector<int> a={1, 2, 3};
std::vector<int> b={4, 5, 6};

a是一个指针指向foo的向量 b是一个指针指向bar

的向量

int* c = a.data();
int* d = b.data();

c是一个指向a指针当前指向的指针; foo
d是一个指针，指向b指针当前指向的内容; bar

现在，让我们做一些交换......

std::swap(c, d);

c是一个指针，现在指向d指向的内容; bar
d是一个指针，现在指向c所指向的内容; foo
因此，c现在指向{4, 5, 6}，而d现在将指向{1, 2, 3}

std::swap(a, b);

a的指针现在指向b指针指向的内容; bar
b的指针现在指向a指针指向的内容; foo
交换向量交换他们指向的内容;它不会交换里面的实际数据。

理解std :: swap的指针和std :: vectors

2 个答案: