有没有一种方法可以避免调用构造函数和析构函数来完美地转发给函数？

Question

我试图将文字传递给完美的转发设置器，但这会导致对象构造，移动和破坏。我正在寻找一种更有效的方法。

struct TStruct {
    TStruct(int va) : a(va) {
        std::cout << "- TStruct(" << va << ")" << std::endl;
    }
    TStruct(const TStruct& other) :
        a(other.a)
    {
        std::cout << "- TStruct(const TStruct& )" << std::endl;
    }
    TStruct(TStruct&& other) :
        a(std::exchange(other.a, 0))
    {
        std::cout << "- TStruct(const TStruct&&)" << std::endl;
    }
    TStruct& operator=(const TStruct& rhs) {
        std::cout << "- TStruct operator= const& " << std::endl;
        // check for self-assignment
        if(&rhs == this) return *this;
        a = rhs.a;
        return *this;
    }
    TStruct& operator=(TStruct&& rhs) {
        std::cout << "- TStruct operator=&&" << std::endl;
        // check for self-assignment
        if(&rhs == this) return *this;
        a = std::exchange(rhs.a, 0);
        return *this;
    }
    ~TStruct() {
        std::cout << "~ TStruct() destructor with " << a << std::endl;
    }
    int a = 1;
};

struct TPerfectForward {
    template<class T>
    TPerfectForward(T&& tsv) : ts(std::forward<T>(tsv)) {}

    template<class T>
    void set(T&& tsv) { ts = std::forward<T>(tsv); }

    TStruct ts;
};

std::cout << "TPerfectForward (5)" << std::endl;
TPerfectForward pf(5);

std::cout << "TPerfectForward set(4)" << std::endl;
pf.set(4);

在Ubuntu上使用gcc 7.4.0可获得以下结果：

TPerfectForward (5);
- TStruct(5)
TPerfectForward set(4)
- TStruct(4)
- TStruct operator=&&
~ TStruct() destructor with 0

我希望这些结果更好：

TPerfectForward (5)
- TStruct(5)
TPerfectForward set(4)
- TStruct(4)

当完美转发到函数时，是否有一种方法可以避免调用move运算符和析构函数？