Prevent expression templates binding to rvalue references

老子叫甜甜 提交于 2019-12-04 02:54:29

Is there any way of designing expression templates to prevent the compiler from compiling such code that may result in the use of expired temporaries at runtime?

No. This was actually recognized before C++11's final standardization, but I don't know if it was ever brought to the committee's notice. Not that a fix would have been easy. I suppose the simplest thing would be a flag on types that would simply error if auto tries to deduce it, but even that would be complex because decltype can also deduce it, as well as template argument deduction. And all three of these are defined in the same way, but you probably don't want the latter to fail.

Just document your library appropriately and hope that nobody tries to capture them that way.

As I understnad, root of your problem is that expression template temporary may have references/pointers to some another temporaries. And by using auto&& we only extend life of expression template temporary itself, but not lifetime of temporaries it has references to. Is it right?

For instance, is this your case?

#include <iostream>
#include <deque>
#include <algorithm>
#include <utility>
#include <memory>
using namespace std;

deque<bool> pool;

class ExpressionTemp;
class Scalar
{
    bool *alive;

    friend class ExpressionTemp;

    Scalar(const Scalar&);
    Scalar &operator=(const Scalar&);
    Scalar &operator=(Scalar&&);
public:
    Scalar()
    {
        pool.push_back(true);
        alive=&pool.back();
    }
    Scalar(Scalar &&rhs)
        : alive(0)
    {
        swap(alive,rhs.alive);
    }
    ~Scalar()
    {
        if(alive)
            (*alive)=false;
    }
};
class ExpressionTemp
{
    bool *operand_alive;
public:
    ExpressionTemp(const Scalar &s)
        : operand_alive(s.alive)
    {
    }
    void do_job()
    {
      if(*operand_alive)
          cout << "captured operand is alive" << endl;
      else
          cout << "captured operand is DEAD!" << endl;
    }
};

ExpressionTemp expression(const Scalar &s)
{
    return {s};
}
int main()
{
    {
        expression(Scalar()).do_job(); // OK
    }
    {
        Scalar lv;
        auto &&rvref=expression(lv);
        rvref.do_job(); // OK, lv is still alive
    }
    {
        auto &&rvref=expression(Scalar());
        rvref.do_job(); // referencing to dead temporary
    }
    return 0;
}

If yes then one of possible solutions, is to make special kind of expression template temporaries which hold resources moved from temporaries.

For instance, check this approach (you may define BUG_CASE macro, to get again bug case).

//#define BUG_CASE

#include <iostream>
#include <deque>
#include <algorithm>
#include <utility>
#include <memory>
using namespace std;

deque<bool> pool;

class ExpressionTemp;
class Scalar
{
    bool *alive;

    friend class ExpressionTemp;

    Scalar(const Scalar&);
    Scalar &operator=(const Scalar&);
    Scalar &operator=(Scalar&&);
public:
    Scalar()
    {
        pool.push_back(true);
        alive=&pool.back();
    }
    Scalar(Scalar &&rhs)
        : alive(0)
    {
        swap(alive,rhs.alive);
    }
    ~Scalar()
    {
        if(alive)
            (*alive)=false;
    }
};
class ExpressionTemp
{
#ifndef BUG_CASE
    unique_ptr<Scalar> resource; // can be in separate type
#endif
    bool *operand_alive;
public:
    ExpressionTemp(const Scalar &s)
        : operand_alive(s.alive)
    {
    }
#ifndef BUG_CASE
    ExpressionTemp(Scalar &&s)
        : resource(new Scalar(move(s))), operand_alive(resource->alive)
    {
    }
#endif
    void do_job()
    {
      if(*operand_alive)
          cout << "captured operand is alive" << endl;
      else
          cout << "captured operand is DEAD!" << endl;
    }
};

template<typename T>
ExpressionTemp expression(T &&s)
{
    return {forward<T>(s)};
}
int main()
{
    {
        expression(Scalar()).do_job(); // OK, Scalar is moved to temporary
    }
    {
        Scalar lv;
        auto &&rvref=expression(lv);
        rvref.do_job(); // OK, lv is still alive
    }
    {
        auto &&rvref=expression(Scalar());
        rvref.do_job(); // OK, Scalar is moved into rvref
    }
    return 0;
}

Your operator/function overloads may return different types, depending on T&&/const T& arguments:

#include <iostream>
#include <ostream>
using namespace std;

int test(int&&)
{
    return 1;
}
double test(const int&)
{
    return 2.5;
};

int main()
{
    int t;
    cout << test(t) << endl;
    cout << test(0) << endl;
    return 0;
}

So, when your expression template temporary do not have resources moved from temporaries - it's size will be not affected.

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