【c++基础】081实际意义的析构函数

#include <cstring>
using namespace std;
static char *num1[] = {"", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};

static char *num10[] = {"", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};

class 
{
private:
    char name[20]; 
    char type[20]; //机器人型号
    int num;       //待翻译的整数
    char *ps;      //指向英文字符串
public:
    robot() //构造函数
    {
        strcpy(name, "XXXXXX");
        strcpy(type, "YYYYYY");
        num = 0;
        ps = new char[5];
        strcpy(ps, "zero");
    }
    void set(char n[], char t[], int m); //设置修改数据
    char *out(int a);                    //英文中每三位数读法相同，所以定义out函数翻译小于1000的整数
    char *tran_int(int n);               //将1~1999999999的整数翻译成英文句子
    void print_num();                    //输出整数及其英文句子
    ~robot() { delete[] ps; };           //析构函数，释放构造函数和set函数中动态申请的ps数组
};
char *robot::out(int a)
{
    char k[1000] = "";
    int b = a % 100;
    //若百位不为0，输出百位数加hundred，若此时十位各位均为0，不加and
    if (a / 100 != 0)
    {
        strcat(k, num1[a / 100]);
        strcat(k, " ");
        strcat(k, "hundred ");
        if (b != 0)
            strcat(k, "and ");
    }
    //当后两位在20以内时，直接调用num1[n]，输出
    if (b < 20)
    {
        strcat(k, num1[b]);
        strcat(k, " ");
    }
    else
    {
        //先调用num10，输出十位数
        strcat(k, num10[b / 10]);
        //个位不为0时应输出“-”个位数
        if (b % 10 != 0)
        {
            strcat(k, "b-");
            strcat(k, num1[b % 10]);
            strcat(k, " ");
        }
    }
    char *p = new char[strlen(k) + 1];
    strcpy(p, k);
    return p;
}
char *robot::tran_int(int n)
{
    char *p;
    char kk[1000] = "";
    if (n > 1999999999)
        strcpy(kk, "数字过大，无法处理！");
    else
    {
        //三位三位取出，存入abcd中
        int a = n / 1000000000, b = (n % 1000000000) / 1000000, c = (n % 1000000) / 1000, d = n % 1000;
        //不等于0时，输出并加上million或thousand
        if (a != 0)
        {
            p = out(a);
            strcpy(kk, p);
            strcat(kk, "billion ");
            delete[] p; //释放在out函数中动态申请的空间
        }
        if (b != 0)
        {
            p = out(b);
            strcpy(kk, p);
            strcat(kk, "million ");
            delete[] p; //释放在out函数中动态申请的空间
        }
        if (c != 0)
        {
            p = out(c);
            strcpy(kk, p);
            strcat(kk, "thousand ");
            delete[] p; //释放在out函数中动态申请的空间
        }
        if (d != 0)
        {
            //根据英文法规则，最后两位前一定有and
            if (d < 100 && (a != 0 || b != 0 || c != 0))
                strcat(kk, "and ");
            p = out(d);
            strcat(kk, p);
            delete[] p;
        }
    }
    p = new char[strlen(kk) + 1];
    strcpy(p, kk);
    return p;
}
void robot::set(char n[], char t[], int m)
{
    strcpy(name, n);
    strcpy(type, t);
    if (num == m)
        return;
    else
    {
        num = m;
        delete[] ps;
    }
    if (num > 0)
    {
        char *tp = tran_int(num);
        ps = new char[strlen(tp) + 1];
        strcpy(ps, tp);
        delete[] tp;
    }
    else if (num == 0)
    {
        ps = new char[5];
        strcpy(ps, "zero");
    }
    else
    {
        ps = new char[13];
        strcpy(ps, "负数不能翻译");
    }
}
void robot::print_num()
{
    cout << ps << endl;
}
int main()
{
    robot brown;
    brown.print_num();
    int n;
    cout << "请输入n：";
    cin >> n;
    brown.set("brown", "800#", n);
    brown.print_num();
    return 0;
}