【More Effective C++】条款5：警惕隐式类型转换

隐式类型转换操作符，可能导致错误的函数调用，应该提供显示的函数调用

class Rational {
public:
    Rational(int numerator = 0, int denominator = 1) : _numerator(numerator), _denominator(denominator) {}
    // 隐式类型转换操作符
    // operator double() const {
    //     return double(_numerator)/_denominator;
    // }
    double asDouble() const {
        return double(_numerator)/_denominator;
    }

private:
    int _numerator;
    int _denominator;
};

int main() {
    Rational r(1, 2);
    std::cout << r << "\n"; // error: no type named ‘type’ in ‘struct std::enable_if<false, std::basic_ostream<char>&>’
    std::cout << r.asDouble() << "\n";
    // double d = r*0.5;
    // std::cout << d << "\n";
    return 0;
}

但个参数的构造函数也会存在隐式转换，考虑如下场景：

• if (a == b[i]) 忘记写了a[i]，编译器不会报错，是因为发生的隐式转换；

  • 转换成这种形式 if (a == static_cast<Array<int>>(b[i]))

• 通过explicit禁止隐式转换，报错信息如下：
  •  error: no match for ‘operator==’ (operand types are ‘Array<int>’ and ‘int’)
• 通过explicit声明的单参数构造函数，还是可以显式调用成功；

template<typename T>
class Array {
public:
    Array(int lowBound, int highBound) {}
    explicit Array(int size) : _size(size) {
        data = new T(_size);
    }
    T& operator[](int index) const {
        if (index >= _size) std::out_of_range("out_of_range");
        return data[index];
    }
    int size() const { return _size; }
private:
    T* data;
    int _size;
};
bool operator==(const Array<int>& lhs, const Array<int>& rhs) {
    if (lhs.size() != rhs.size()) return false;
    for (int i = 0; i < lhs.size(); i++) {
        if (lhs[i] != rhs[i]) return false;
    }
    return true;
}

int main() {
    Array<int> a(2); a[0] = 1; a[1] = 2;
    Array<int> b(2); b[0] = 1; b[1] = 2;
    for (int i = 0; i < 2; i++) {
        // if (a == b[i]) {
        if (a == static_cast<Array<int>>(b[i]))
            std::cout << "equal\n";
        } else {
            std::cout << "not equal\n";
        }
    }
    return 0;
}

还可以通过proxy class方式来禁止隐式转换：

• if (a == b[i])，该步骤中隐式转换会经过两次用户定制的转换，该行为是不被允许的
  • b[i]从一个int先会被转换为ArraySize类型，
  • 然后从ArraySize转换为Array<int>类型。

template<typename T>
class Array {
public:
    class ArraySize {
    public:
        ArraySize(int numElements) : theSize(numElements) {}
        int size() const { return theSize; }
    private:
        int theSize;
    };
    Array(ArraySize size) : _size(size.size()) {
        data = new T(_size);
    }
    T& operator[](int index) const {
        if (index >= _size) std::out_of_range("out_of_range");
        return data[index];
    }
    int size() const { return _size; }
private:
    T* data;
    int _size;
};