Question

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

隐式转换在模板参数推导期间不适用，您可以渲染函数friend（以便知道类型）：

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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