数组元素ch[j]=='('报错，我是想作比较不是赋值，编译器报错->'=' : left operand must be l-value 。

Question

数组元素ch[j]=='('报错，我是想作比较不是赋值，编译器报错->'=' : left operand must be l-value （请帮忙看看哪里有问题，附上源码和报错）。

源码:

#include <iostream>
#include <string.h>
// #include <stdio.h>
// #include <cmath>
#include <string>
using namespace std;

class Rational
{
protected:
    string str;
    int integer;  //整数部分
    int point[2]; //小数部分，又分为非循环与循环部分

public:
    Rational();
    Rational(string st);
    void countInteger();         //取整数部分
    void countPoint();           //取小数部分，计算作为小数部分时分母和分子
    int countSimplify();         //当只有小数部分时，化简
    void countAdd();             //，化简后整合两部分，相加
    void integerTransfDecimal(); //show
    friend istream &operator>>(istream &input, Rational &Ra);
};

Rational::Rational()
{
    this->str = "0";
    this->integer = 0;
    this->point[0] = 0;
    this->point[1] = 0;
}

Rational::Rational(string st)
{
    this->str = st;
    this->integer = 0;
    this->point[0] = 0;
    this->point[1] = 0;
}
//取整数部分
void Rational::countInteger()
{
    char ch[15];
    for (int i = 0; i < str.length(); i++)
    {
        if (ch[i] != '.')
        {
            ch[i] = str[i];
            this->point[0]++; //临时计数
        }
        else
            break;
    }
}
//取小数部分，计算作为小数部分时分母和分子
void Rational::countPoint()
{
    char ch[15];
    int te = this->point[0];
    for (int i = te + 1; i < str.length(); i++)
    {
        ch[i] = str[i];
    }
    int px = 0, py = 0, pm = 0;
    //计算位数
    for (int j = 0; j < str.size(); j++)
    {
        if (ch[j] != '(' &&pm = 0)
        {
            this->point[0] *= 10;
            this->point[0] += ch[j] - '0'; //得到非循环数值
            px++;                          //记非循环位数
        }
        if (ch[j] == '(' &&pm = 0)
        {
            pm = 1;
            j++; //记循环内位数
        }
        if (ch[j] != ')' &&pm = 1)
        {
            this->point[1] *= 10;
            this->point[1] += ch[j] - '0';
            py++;
        }
    }

    //判断类型，计算分母分子最终值的大小
    if (pm == 1) //如果为非循环小数(例:0.5(0))
    {
        int pw = px - 1;
        int pn = this->point[0];
        this->point[1] = 9;
        while (pw--)
        {
            this->point[0] *= 10;
        }
        pw = px;
        while (pw--)
        {
            this->point[1] *= 10;
        }
        this->point[0] -= pn;
    }
    if (pm == 0 && px == 1) //如果为纯循环小数(例:0.(3))
    {
        int pw = py;
        int pn = this->point[0];
        this->point[1] = 9;
        while (pw--)
        {
            this->point[0] *= 10;
            this->point[1] *= 10;
            this->point[1] += 9;
        }
        this->point[0] -= pn;
    }
    if (pm == 0 && px == 0) //如果为非纯循环小数(例:0.52(3))
    {
        int pw = px;
        int pu = py;
        int pn = this->point[0];
        this->point[1] = 9;
        while (pu--)
        {
            this->point[0] *= 10;
        }
        while (pw--)
        {
            this->point[1] *= 10;
        }
        this->point[0] -= pn;
    }
}

//当只有小数部分时，化简
int Rational::countSimplify()
{
    int te = 0;
    for (int i = 1; i <= this->point[1]; i++)
    {
        if (this->point[0] / i == this->point[1] / i)
        {
            te = i; //i最小值应该为1
        }
    }
    return te;
}

//，化简后整合两部分，相加
void Rational::countAdd()
{
    this->integer *= this->point[1];
    this->point[0] *= this->integer;
}
//show
void Rational::integerTransfDecimal()
{
    this->countInteger();
    this->countPoint();
    this->countSimplify();
    this->countAdd();
    cout << "\ninteger transform decimal is :" << endl;
}

istream &operator>>(istream &input, Rational &Ra)
{
    cout << "Please input a point(小数)" << endl;
    input >> Ra.str;
    return input;
}

int main()
{
    Rational ra1;
    cin >> ra1;
    ra1.integerTransfDecimal();
    return 0;

编译错误提示:

: error C2106: '=' : left operand must be l-value
: error C2106: '=' : left operand must be l-value
: error C2106: '=' : left operand must be l-value
 : error C2248: 'str' : cannot access protected member declared in class 'Rational'
        : see declaration of 'str'
 : error C2593: 'operator >>' is ambiguous
执行 cl.exe 时出错.

测试八.exe - 1 error(s), 0 warning(s)

谢谢谢谢!

Answer 1

if (ch[j] != '(' &&pm = 0)
->
if (ch[j] != '(' &&pm == 0)

istream &operator>>(istream &input, Rational &Ra)
这个要么申明成友元函数，要么把str定义为public