具有Overload []运算符的C ++ MyInteger类，因此索引在位置i返回数字

Question

The problem I have

Define a class named MyInteger that stores an integer and has functions to get and set the intergers value. Then, overload the [] operator so that the index returns the digit in position i, where i = 0 is the least-significant digit. If no such digit exists then -1 should be returned.

I understand the first part but I don't understand how to implement the second part,

Then, overload the [] operator so that the index returns the digit in position i, where i = 0 is the least-significant digit. If no such digit exists then -1 should be returned.

My code is such:

#include <iostream>
using namespace std;

class MyInteger{
private:
  int integer;
public:
  MyInteger(int bInteger = 0)
  {
     integer = bInteger;
  }
  int getInteger(){ return integer; }
  void setInteger(int bInteger){ integer = bInteger; }

  void operator[](int x) //???
  {
  }

};

int main()
{

  cout << "Enter an integer " << endl;
  //cin >> 

}

Answer 1

As i understand the second part you have to implement the operator to return a single digit. In a way that given

MyInteger x;
x.set(43210)
int a = x[0];
int b = x[1];
int c = x[5];

a should equal 0, b should equal 1 and c should equal -1. To get the digits you could use modulo arithmetic:

To get the first digit (index 0) of any integer "i" just return i%10. To get the second digit (index 1)

int result = ((i - (i%10)) % 100) / 100

for the 2nd index

int result = ((i - (i%10) - (i%100)) % 1000) / 1000

This may be formulated more elegently with a for loop.

Answer 2

对于十进制数字，您的类应返回与位置（索引）相对应的十进制数字，例如，如果MyInteger m = 12347，则运算符m [0]将返回7，m [3]将返回2，而m [6]将返回-1。

Answer 3

For your first part of the question, I propose you the following:

int operator[](int x) 
{
    int ten; 
    for (ten = 10; x; x--)
        ten *= 10; 
    int digit = (integer%ten) / (ten / 10);
    return digit; 
}

By the way, to ease tests, I propose you to define the following to permit using cin on MyInteger :

istream& operator>> (istream& is, MyInteger& x) 
{
    int v; 
    is >> v;
    x.setInteger(v); 
    return is;
} 

For the second part , If no such digit exists then -1 should be returned. , addresses the "out of bound" that you could have. For example, if you'd have x = 21 and you'd do x[3], you'd return 0 with the programme above whereas in reality there is no fourth digit.

So you could change the last instruction of operator[] as follows:

    return integer>=ten/10 ?  digit:-1; 

Answer 4

I can share you my code. But i not sure if it is best to solve your problem.

 #include<iostream> #include<cmath> using namespace std; class MyInteger { public: MyInteger(int integer); MyInteger(); int operator[](int index); void set(int integer); int get(); private: int integer; }; int main() { MyInteger integer; integer.set(123456); cout<<"MyInteger: "<<integer.get()<<endl; int i; /*position*/ cout<<"Enter the reference from zero to fine digit>>"; cin>>i; if(integer[i]==-1) cout<<"The integer doesn't exit.\\n"; else cout<<"The digit is: "<<integer[i]; system("pause"); return 0; } MyInteger::MyInteger(int integer) { this->integer=integer; } MyInteger::MyInteger() { this->integer=0; } int MyInteger::operator[](int index) { /*檢測是否在範圍內*/ int size=0; int replacement=this->integer; do { replacement=replacement/10; size++; }while(replacement!=0); if(index>=size) { cout<<"It is not in range!\\n"; return -1; } /*找出位置上的digit*/ int target=0; int head=pow(10.0,index+1); /*去頭用*/ target=(this->integer)-((this->integer)-(this->integer)%head); int foot=pow(10.0,index); /*去尾用*/ target=target/foot; return target; } void MyInteger::set(int integer) { if((integer-(int)integer)==0) this->integer=integer; else { cout<<"It is not a intger!!"; exit(1); } } int MyInteger::get() { return this->integer; }