C ++中的字符串和整数

Question

First off, this is for a homework assignment, so I'd appreciate help and guidance rather than just the answer in code.

The purpose of the code should be for a user to input a number and a width.

If the width is longer than the number, the number will be printed out with zeros in front of the number. For example 43 3 would give 043 .

If the width isn't longer just the number would be printed: 433 2 would be 433 .

I think I have to get the count of characters in the number and compare it to the count of characters in the width ( if-else statement).

Then, if the number of characters in the number is more, print out the number. Else, print out the width.

I think I get the number of zeros by subtracting the length of the number from the length of the width. Then use that to set the number of zeros. Like I said this is homework and would rather learn than be given the answer.

If anyone can help, it'll be appreciated.

    #include <iostream>;
    #include <string>;

    using namespace std;

    string format(int number, int width) {


    int count = 0;
      if (number > width)// This if-else is incomplete
          return ;  
      else              

    }

    int main() 
    {
     cout << "Enter a number: ";
     string n;
     cin >> n;

     cout << "Enter the number's width: ";
     string w;
     cin >> w;

     format(n, w);

    }

Answer 1

no need to checking string or other things write these code C++ will do it for you automatically.

#include <conio.h>
#include <iostream>
using std::cout;
using std::cin;

#include <string>;
using std::string;

#include <iomanip>
using std::setw;

void format(int number, int width)
{
    cout.fill('0');

    cout << setw(width) << number;
}

int main()
{
    cout << "Enter a number: ";
    int n;
    cin >> n;

    cout << "Enter the number's width: ";
    int w;
    cin >> w;

    format(n, w);

    _getch();
    return 0;
}