c++ - std::move 如何使原始变量的值无效？

Question

在cpp 参考的以下示例中：

#include <iostream>
#include <utility>
#include <vector>
#include <string>

int main()
{
    std::string str = "Hello";
    std::vector<std::string> v;

    // uses the push_back(const T&) overload, which means 
    // we'll incur the cost of copying str
    v.push_back(str);
    std::cout << "After copy, str is \"" << str << "\"\n";

    // uses the rvalue reference push_back(T&&) overload, 
    // which means no strings will be copied; instead, the
    // Contents of str will be moved into the vector.  This is
    // less expensive, but also means str might now be empty.
    v.push_back(std::move(str));
    std::cout << "After move, str is \"" << str << "\"\n";

    std::cout << "The contents of the vector are \"" << v[0]
              << "\", \"" << v[1] << "\"\n";
}

使用std::move可能会导致原始值丢失。对我来说，它看起来像

v.push_back(std::move(str))

导致v[1]创建一个新成员。然后，

&v[1] = &str

但为什么它会损害 中的价值str呢？它没有任何意义。

有许多复杂的教程std::move比我自己的问题更难理解。

有人可以写吗

v.push_back(std::move(str))

在其等效使用c++03？

我寻找一种易于理解且不包含诸如x-value、static_cast、remove_reference等先决条件的解释，因为它们本身需要先理解std::move。请避免这种循环依赖。

这些链接也不能回答我的问题：7510182、3413470

因为我有兴趣知道str伤害是如何产生的，而不是会发生什么v[1]。

伪代码也很受欢迎，只要它像c++03.

---

更新：为避免复杂化，让我们考虑一个更简单的示例，int如下所示

int x = 10;
int y = std::move(x);
std::cout << x;

标签： c++c++11movemove-semanticsstdmove

Answer 1

根据实现，这std::move 可能是内部存储器地址的简单交换。

如果您在http://cpp.sh/9f6ru上运行以下代码

#include <iostream>
#include <string>

int main()
{
  std::string str1 = "test";
  std::string str2 = "test2";

  std::cout << "str1.data() before move: "<< static_cast<const void*>(str1.data()) << std::endl;
  std::cout << "str2.data() before move: "<< static_cast<const void*>(str2.data()) << std::endl;

  str2 = std::move(str1);
  std::cout << "=================================" << std::endl;

  std::cout << "str1.data() after move: " << static_cast<const void*>(str1.data()) << std::endl;
  std::cout << "str2.data() after move: " << static_cast<const void*>(str2.data()) << std::endl;
}

您将获得以下输出：

str1.data() before move: 0x363d0d8
str2.data() before move: 0x363d108
=================================
str1.data() after move: 0x363d108
str2.data() after move: 0x363d0d8

但结果可能会因编译器和 std 库的实现而异。

但实现细节可能更复杂http://cpp.sh/6dx7j。如果您查看您的示例，那么您将看到为字符串创建副本并不一定需要为其内容分配新内存。这是因为几乎所有的操作std::string都是只读的或者需要分配内存。所以实现可以决定只做浅拷贝：

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

int main()
{
  std::string str = "Hello";
  std::vector<std::string> v;

  std::cout << "str.data() before move: "<< static_cast<const void*>(str.data()) << std::endl;

  v.push_back(str);
  std::cout << "============================" << std::endl;
  std::cout << "str.data()  after push_back: "<< static_cast<const void*>(str.data()) << std::endl;
  std::cout << "v[0].data() after push_back: "<< static_cast<const void*>(v[0].data()) << std::endl;

  v.push_back(std::move(str));
  std::cout << "============================" << std::endl;

  std::cout << "str.data()  after move: "<< static_cast<const void*>(str.data()) << std::endl;
  std::cout << "v[0].data() after move: "<< static_cast<const void*>(v[0].data()) << std::endl;
  std::cout << "v[1].data() after move: "<< static_cast<const void*>(v[1].data()) << std::endl;
  std::cout << "After move, str is \"" << str << "\"\n";


  str = std::move(v[1]);
  std::cout << "============================" << std::endl;
  std::cout << "str.data()  after move: "<< static_cast<const void*>(str.data()) << std::endl;
  std::cout << "v[0].data() after move: "<< static_cast<const void*>(v[0].data()) << std::endl;
  std::cout << "v[1].data() after move: "<< static_cast<const void*>(v[1].data()) << std::endl;
  std::cout << "After move, str is \"" << str << "\"\n";
}

输出是

str.data() before move: 0x3ec3048
============================
str.data()  after push_back: 0x3ec3048
v[0].data() after push_back: 0x3ec3048
============================
str.data()  after move: 0x601df8
v[0].data() after move: 0x3ec3048
v[1].data() after move: 0x3ec3048
After move, str is ""
============================
str.data()  after move: 0x3ec3048
v[0].data() after move: 0x3ec3048
v[1].data() after move: 0x601df8
After move, str is "Hello"

如果你看一下：

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

int main()
{
  std::string str = "Hello";
  std::vector<std::string> v;

  std::cout << "str.data() before move: "<< static_cast<const void*>(str.data()) << std::endl;

  v.push_back(str);
  std::cout << "============================" << std::endl;
  str[0] = 't';
  std::cout << "str.data()  after push_back: "<< static_cast<const void*>(str.data()) << std::endl;
  std::cout << "v[0].data() after push_back: "<< static_cast<const void*>(v[0].data()) << std::endl;

}

然后你会假设这str[0] = 't' 只会替换数据。但这不一定是http://cpp.sh/47nsy的情况。

str.data() before move: 0x40b8258
============================
str.data()  after push_back: 0x40b82a8
v[0].data() after push_back: 0x40b8258

并移动原语，例如：

void test(int i) {
  int x=i;
  int y=std::move(x);
  std::cout<<x;
  std::cout<<y;
}

大部分将由编译器完全优化：

  mov ebx, edi
  mov edi, offset std::cout
  mov esi, ebx
  call std::basic_ostream<char, std::char_traits<char> >::operator<<(int)
  mov edi, offset std::cout
  mov esi, ebx
  pop rbx
  jmp std::basic_ostream<char, std::char_traits<char> >::operator<<(int) # TAILCALL

两者都std::cout 使用相同的寄存器，x并且y完全优化了。