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std::move如何使原始变量的值无效?

stdmove move-semantics move c++11 c++

-2

ar2015 · 技术社区 · 6 年前

在下面的示例中 cpp reference :

#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";
}

std::move 可能导致原值丢失。在我看来

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

v[1] 被创造。那么,

&v[1] = &str

但为什么它会损害 str ? 这没有道理。

有很多复杂的教程这比我自己的问题更难理解。

有人能写信吗

v、 向后推(std::move(str))

等同于 c++03 ?

我寻找一个解释,其理解是容易的,不包含诸如 x-value , static_cast remove_reference 等等,因为他们自己需要理解第一。请避免这种循环依赖关系。

7510182 , 3413470

因为我想知道 str公司 是伤害而不是伤害 五[1] .

c++03语言

更新:为了避免复杂化,让我们考虑一个更简单的例子 int 如下所示

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

2 回复 | 直到 6 年前

t.niese 6 年前

根据实现的不同 std::move 可能是内部内存地址的简单交换。

http://cpp.sh/9f6ru

#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;
}

您将得到以下输出:

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

但是实现细节可能更复杂 http://cpp.sh/6dx7j . 如果您查看您的示例,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 是只读的或需要分配内存。因此,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";
}

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"

如果你看看:

#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;

}

那么你会认为 str[0] = 't' 只是把数据放回原处。但事实并非如此 http://cpp.sh/47nsy .

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

移动原语,比如:

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

主要是由编译器完全优化出来的:

  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

std::cout 使用相同的寄存器 x 和 y

Deduplicator 6 年前

std::move 是对rvalue引用的简单强制转换。实际上不是做

所有的魔法都发生在函数中接收 掠夺许可证

因此,使用移动语义通常效率更高(一个人可以做得更快,相信我),并且不太可能抛出异常(资源获取容易失败),因为这样做的代价是破坏源代码。

所有这些都是由一个叫做 标准::移动 ,但它本身做