CODEWITHSUNDEEP
c++11movemove-semantics
dani
dani

Reputation: 3887

Is there a way to make a moved object "invalid"?

I've some code that moves an object into another object. I won't need the original, moved object anymore in the upper level. Thus move is the right choice I think.

However, thinking about safety I wonder if there is a way to invalidate the moved object and thus preventing undefined behaviour if someone accesses it.

Here is a nice example:

// move example
#include <utility>      // std::move
#include <vector>       // std::vector
#include <string>       // std::string

int main () {
  std::string foo = "foo-string";
  std::string bar = "bar-string";
  std::vector<std::string> myvector;

  myvector.push_back (foo);                    // copies
  myvector.push_back (std::move(bar));         // moves

  return 0;
}

The description says:

The first call to myvector.push_back copies the value of foo into the vector (foo keeps the value it had before the call). The second call moves the value of bar into the vector. This transfers its content into the vector (while bar loses its value, and now is in a valid but unspecified state).

Is there a way to invalidate bar, such that access to it will cause a compiler error? Something like:

myvector.push_back (std::move(bar));         // moves
invalidate(bar); //something like bar.end() will then result in a compiler error

Edit: And if there is no such thing, why?

Upvotes: 2

Views: 437

Answers (3)

Johannes Schaub - litb
Johannes Schaub - litb

Reputation: 507055

The bigger picture answer is because moving or not moving is a decision made at runtime, and giving a compile-time error is a decision made at compile time. 

foo(bar); // foo might move or not
bar.baz(); // compile time error or not?

It's not going to work.. you can approximate in compile time analysis, but then it's going to be really difficult for developers to either not get an error or making anything useful in order to keep a valid program or the developer has to make annoying and fragile annotations on functions called to promise not to move the argument.

To put it a different way, you are asking about having a compile time error if you use an integer variable that contains the value 42. Or if you use a pointer that contains a null pointer value. You might be succcessful in implementing an approximate build-time code convention checker using clang the analysis API, however, working on the CFG of the C++ AST and erroring out if you can't prove that std::move has not been called till a given use of a variable.

Upvotes: 1

Lapshin Dmitry
Lapshin Dmitry

Reputation: 1124

Move semantics works like that so you get an object in any it's correct state. Correct state means that all fields have correct value, and all internal invariants are still good. That was done because after move you don't actually care about contents of moved object, but stuff like resource management, assignments and destructors should work OK. All STL classes (and all classed with default move constructor/assignment) just swap it's content with new one, so both states are correct, and it's very easy to implement, fast, and convinient enough.

You can define your class that has isValid field that's generally true and on move (i. e. in move constructor / move assignment) sets that to false. Then your object will have correct state I am invalid. Just don't forget to check it where needed (destructor, assignment etc).

That isValid field can be either one pointer having null value. The point is: you know, that object is in predictable state after move, not just random bytes in memory.

Edit: example of String:

class String {
public:
    string data;
private:
    bool m_isValid;

public:
    String(string const& b): data(b.data), isValid(true) {}

    String(String &&b): data(move(b.data)) {
        b.m_isValid = false;
    }

    String const& operator =(String &&b) {
        data = move(b.data);
        b.m_isValid = false;
        return &this;
    }

    bool isValid() {
        return m_isValid;
    }
}

Upvotes: 0

user1084944
user1084944

Reputation:

Accessing the moved object is not undefined behavior. The moved object is still a valid object, and the program may very well want to continue using said object. For example,

template< typename T >
void swap_by_move(T &a, T &b)
{
    using std::move;
    T c = move(b);
    b = move(a);
    a = move(c);
}

Upvotes: 1

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