std :: move如何使原始变量的值无效? [英] How does std::move invalidates the value of original variable?

问题描述

在来自 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.在我看来，

Using std::move may cause the original value be lost. To me, it looks like

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

将导致创建新成员v[1].然后，

causes a new member v[1] being created. Then,

&v[1] = &str

但是为什么要破坏str中的值?这没有道理.

But why should it damage the value in str? It does not make sense.

关于std::move的教程很复杂，比我自己的问题难理解.

There are many complicated tutorials about std::move which are harder than my own question to understand.

任何人都可以写

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

等同于使用c++03吗?

我正在寻找一个易于理解的解释，并且不包含诸如x-value，static_cast和remove_reference等先决条件，因为它们本身首先需要理解std::move.请避免这种循环依赖.

I look for an explanation whose understanding is easy and do not contain prerequisites such as x-value , static_cast, remove_reference, etc, as they themselves require to understand std::move first. Please avoid this circular dependency.

这些链接也无法回答我的问题: 7510182 ， 3413470

Also these links do not answer my question: 7510182, 3413470

因为我有兴趣了解str的危害，而不是v[1]的危害.

Because I am interested in knowing how str is harm and not what happens to v[1].

只要是c++03一样简单，也欢迎使用伪代码.

Pseudo code is also welcome as far as it is as simple as c++03.

更新:为避免复杂起见，让我们考虑一个更简单的int示例，如下所示

Update: To avoid complication, let's consider a simpler example of int as follows

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

推荐答案

根据实现的不同，std::move可以是内部存储器地址的简单交换.

Depending on the implementation, the std::move could be a simple swap of the internal memory addresses.

如果您在 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

但是结果可能会有所不同，具体取决于编译器和std库的实现.

But the result may vary depending on the implementation of the compiler and the std library.

但是实现细节可能更加复杂 http://cpp.sh/6dx7j .如果看一下示例，您将看到为字符串创建副本不一定需要为其内容分配新的内存.这是因为std::string上的几乎所有操作都是只读的，或者需要分配内存.因此，实现可以决定只做浅表副本:

But the implementation details can be even more complex http://cpp.sh/6dx7j. If you look at your example, then you will see that creating a copy for a string does not necessarily require that new memory for its content is allocated. This is because nearly all operations on std::string are read only or require the allocation of memory. So the implementation can decide to do just shallow copies:

#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";
}

输出为

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"

如果您看一下:

#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，您将假定str[0] = 't'只是替换就位的数据.但这不一定是 http://cpp.sh/47nsy .

Then you would assume that str[0] = 't' would just replace the data in place. But this is not necessarily the case http://cpp.sh/47nsy.

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

移动基本体，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:

Would be mostly be optimized out completely by the compiler:

  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完全被优化.

Both std::cout used the same register, the x and y are completely optimized away.

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