类与封装形态

Posted 2021-09-04 阿弥陀佛.a

一个类可以有很多对象，而一个对象必然属于某个类

#include <stdio.h>

int i = 1;

struct Test
{
private:
    int i;

public:
    int j;
        
    int getI()
    {
        i = 3;
        
        return i;
    }
};

int main()
{
    int i = 2;
    
    Test test;
    
    test.j = 4;
    
    printf("i = %d\\n", i);              // i = 2;
    printf("::i = %d\\n", ::i);          // ::i = 1;默认命名空间的i
    // printf("test.i = %d\\n", test.i);    // Error
    printf("test.j = %d\\n", test.j);    // test.j = 4
    printf("test.getI() = %d\\n", test.getI());  // test.getI() = 3
    
    return 0;
}

//.h
#ifndef _OPERATOR_H_
#define _OPERATOR_H_

class Operator
{
private:
    char mOp;
    double mP1;
    double mP2;
    
public:
    bool setOperator(char op);//设置操作符
    void setParameter(double p1, double p2);//设置操作数
    bool result(double& r);//是否计算成功
};
#endif

//.cpp
#include "Operator.h"

bool Operator::setOperator(char op)
{
    bool ret = false;
        
    if( (op == '+') || (op == '-') || (op == '*') || (op == '/') )
    {
        ret = true;
        mOp = op;
    }
    else
    {
        mOp = '\\0';
    }
        
    return ret;
}

void Operator::setParameter(double p1, double p2)
{
    mP1 = p1;
    mP2 = p2;
}
    
bool Operator::result(double& r)
{
    bool ret = true;
        
    switch( mOp )
    {
        case '/':
            if( (-0.000000001 < mP2) && (mP2 < 0.000000001) )
            {
                ret = false;
            }
            else
            {
                r = mP1 / mP2;
            }
            break;
        case '+':
            r = mP1 + mP2;
            break;
        case '*':
            r = mP1 * mP2;
            break;
        case '-':
            r = mP1 - mP2;
            break;
        default:
            ret = false;
            break;
    }
        
    return ret;
}

//main.cpp
#include <stdio.h>
#include "Operator.h"

int main()
{
    Operator op;
    double r = 0;
    
    op.setOperator('/');
    op.setParameter(9, 3);
    
    if( op.result(r) )
    {
        printf("r = %lf\\n", r);
    }
    else
    {
        printf("Calculate error!\\n");
    }
    return 0;
}