C++:在模板中隐式转换,为什么它不起作用？

Question

这看起来很奇怪,但这个简单的代码使用int而不是T,并且不适用于模板T.

template <typename T>
class Polynomial {

public:
    Polynomial (T i) {}
    Polynomial& operator+= (const Polynomial& rhs) {
        return *this;
    }
};

template <typename T>
const Polynomial<T> operator+ (Polynomial<T> lhs_copy, const Polynomial<T>& rhs) {
    return lhs_copy += rhs;
}

Polynomial<int> x (1), y = x + 2; // no match for 'operator+' in 'x + 2'

Answer 1

在模板参数推导期间,隐式转换不适用,您可能会渲染您的函数friend(以便类型已知):

template <typename T>
class Polynomial {
public:
    Polynomial (T i) {};
    Polynomial& operator+= (const Polynomial& rhs) { return *this; };

    friend Polynomial operator+ (Polynomial lhs, const Polynomial& rhs) {
        return lhs+=rhs;
    }
};

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