Question

在下面的示例中，如何引用当前对象实例以获取使用输出重载的机会？

class Shape {
    private:
        double _length, _width;
        double straight(double value) {
            if (value<0) { return -value; }
            if (value==0) { return 1; }
            return value;
        }
    public:
        Shape() { setDims(1,1); }
        Shape(double length, double width) {
            setDims(length, width); }
        void setDims(double length, double width) {
            _length=straight(length); _width=straight(width); }

        friend ostream &operator<<(ostream &output, Shape &S) {
            output << S._length << "," << S._width; return output; }

        void display() { cout << [THIS] << endl; }
};

int main(int argc, const char * argv[]) {

    Shape s1; s1.display();
    return 0;
}

Answer 1

就像这样：

void display() { cout << *this << endl; }

Answer 2

this是一个指针。您的operator<<想要一个实际的Shape对象，而不是指针。

所以你必须先取消引用指针：*this。

Answer 3

或者只使用运算符＆lt;＆lt;

#include <iostream>

using namespace std;

class Shape {
private:
    double _length, _width;
    double straight(double value) {
        if (value<0) { return -value; }
        if (value == 0) { return 1; }
        return value;
    }

public:
    Shape() { setDims(1, 1); }
    Shape(double length, double width) {
        setDims(length, width);
    }
    void setDims(double length, double width) {
        _length = straight(length); _width = straight(width);
    }

friend ostream &operator<<(ostream &output, Shape &S) {
    output << S._length << "," << S._width; return output;
}

int main(int argc, const char * argv[]) {

    Shape s1;

    std::cout << s1 << std::endl;
}

如何cout＆＃39;这个＆＃39;超载输出？

3 个答案: