问题
In the following examples from 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";
}
Using std::move may cause the original value be lost. To me, it looks like
v.push_back(std::move(str))
causes a new member v[1] being created. Then,
&v[1] = &str
But why should it damage the value in str? It does not make sense.
There are many complicated tutorials about std::move which are harder than my own question to understand.
Could any one please write
v.push_back(std::move(str))
in its equivalent using c++03?
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.
Also these links do not answer my question: 7510182, 3413470
Because I am interested in knowing how str is harm and not what happens to v[1].
Pseudo code is also welcome as far as it is as simple as c++03.
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;
回答1:
Depending on the implementation, the std::move could be a simple swap of the internal memory addresses.
If you run the following code on 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;
}
You will get the following output:
str1.data() before move: 0x363d0d8
str2.data() before move: 0x363d108
=================================
str1.data() after move: 0x363d108
str2.data() after move: 0x363d0d8
But the result may vary depending on the implementation of the compiler and the std library.
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";
}
The output is
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"
And if you take a look at:
#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;
}
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
And moving primitives like:
void test(int i) {
int x=i;
int y=std::move(x);
std::cout<<x;
std::cout<<y;
}
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
Both std::cout used the same register, the x and y are completely optimized away.
回答2:
std::move is a simple cast-to-rvalue-reference. It doesn't actually do anything.
All the magic happens in functions receiving such an rvalue-reference, if they accept it as an rvalue-reference. They take it as license to plunder those objects ruthlessly, obviating the need to actually allocate resources and otherwise do the heavy work of making a copy. Swapping source and target to avoid even the need for cleanup is expected when assigning.
Thus, using move-semantics is generally more efficient (one can do less faster, trust me), and less likely to throw exceptions (resource-acquisition is prone to fail), for the cost of trashing the source.
And all of it is enabled by that small disguised cast known as std::move, which doesn't itself do anything.
来源：https://stackoverflow.com/questions/51916526/how-does-stdmove-invalidates-the-value-of-original-variable