Question

我有两个问题：

我正在尝试重载istream运算符＆gt;＆gt;接受“（a，b）”的输入并设置a = real和b = imaginary。我可以跳过第一个（使用input.ignore（），但输入的数字可以是任何大小，所以我不知道如何只读取“a”到逗号到真实，只是b到虚构。
< / LI>
当我尝试设置number.real时，我得到错误成员Complex :: real是不可访问的。

头

class Complex
{
    friend istream &operator>>(istream &, Complex &);
    friend ostream &operator<<(ostream &, const Complex &);

private:
    double real;
    double imaginary;
};

来源

#include <iostream>
#include <string>
#include <iomanip>
#include "complex.h"
using namespace std;

Complex::Complex(double realPart, double imaginaryPart)
:real(realPart),
imaginary(imaginaryPart)
{
} 
istream &operator>>(istream &input, Complex &number)
{
    string str;
    int i = 0;

    input >> str;

    while (str[i] != ','){
        i++;
    }
    input.ignore();                                     // skips the first '('
    input >> setw(i - 1) >> number.real; //??           "Enter a complex number in the form: (a, b)

}

主

#include <iostream>
#include "Complex.h"
using namespace std;

int main()

{
    Complex x, y(4.3, 8.2), z(3.3, 1.1), k;

    cout << "Enter a complex number in the form: (a, b)\n? ";
    cin >> k; // demonstrating overloaded >>
    cout << "x: " << x << "\ny: " << y << "\nz: " << z << "\nk: "
        << k << '\n'; // demonstrating overloaded <<
}

Answer 1

我认为一个答案可以是write a std::streambuf然后是use the std::streambuf in an inherited std::iostream

Answer 2

如果我理解正确，您只需将值直接读入数据成员：

istream& operator>>(istream& input, Complex& number)
{
    input.ignore(); // ignore the first (
    input >> number.real;
    input.ignore(); // ignore the ,
    input >> number.imaginary;
    input.ignore(); // ignore the next )

    return input;
}

重载运算符＆gt;＆gt;和istream格式

2 个答案: