Hecadecimal

[cblue]

Can anyone give me an idea how to convert an int or decimal number into hexadecimal? I know how to do it for binary by dividing by 2 and using the modulus % operator, but how do you take into account a b c d e f if the integer is greater than 9? any ideas would greatly be appreciated.

[Dave Sinkula]

http://www.daniweb.com/code/search.php?query=decimal

[cblue]

thanks guys, now for the challenge, how would i convert an int to hexadecimal, using recursion, in other words write a recursive function, i know how to do it for base 2, but how do you accomdate for the letters a-f if the digits are greater than 9?

[Narue]

>but how do you accomdate for the letters a-f if the digits are greater than 9?
Use a string:

static const string hex ( "0123456789ABCDEF" );

This is a simple problem once you figure it out. The function shouldn't need to be more than 4 or 5 logical lines.

[cblue]

This is what i'm trying to do but am not sure how to convert it if it is > 9 to a b c d e or f?

Here is the code thus far:

#include <iostream>
using namespace std;
#include "prompt.h"

// recursion: print English form of each digit
// in an integer: 123 -> "one two three"

void PrintDigit(int num)
// precondition: 0 <= num < 10
// postcondition: prints english equivalent, e.g., 1->one,...9->nine     
{
    if (0 == num)       cout << "zero";
    else if (1 == num)  cout << "one";
    else if (2 == num)  cout << "two";
    else if (3 == num)  cout << "three";
    else if (4 == num)  cout << "four";
    else if (5 == num)  cout << "five";
    else if (6 == num)  cout << "six";
    else if (7 == num)  cout << "seven";
    else if (8 == num)  cout << "eight";
    else if (9 == num)  cout << "nine";
    else cout << "?";
}

void Print(long int number)
// precondition: 0 <= number
// postcondition: prints English equivalent of number
{
	
	if(0 <= number && number < 16)
	{
		PrintDigit(int(number));
	}
	else
	{
		Print(number /16);
		cout << " ";
		PrintDigit(int(number % 16));
	}
	
		
}

int main()
{
	long number = PromptRange("enter an integer",0L,1000000L);
    Print(number);
    cout << endl;
    
    return 0;
}

not sure how to test if digit > 9 to print the appropriate letter a = 10 b =11
...f = 15.

<< moderator edit: added [code][/code] tags >>

[Narue]

You're still thinking in base ten. Start thinking in base sixteen and things will be easier:

#include <iostream>
#include <string>

using namespace std;

void print_hex ( int dec )
{
  static const string hex_dig ( "0123456789ABCDEF" );

  if ( dec != 0 ) {
    print_hex ( dec / 16 );
    cout<< hex_dig[dec % 16];
  }
}

int main()
{
  cout<<"0x";
  print_hex ( 27 );
  cout<<endl;
}

[cblue]

thanks everyone for your help, igot this one figured out.
you can delete or dis reguard this thread.

[Narue]

>igot this one figured out
I sure hope so, seeing as how I gave you a complete solution.

>you can delete or dis reguard this thread
Threads don't get deleted when the problem is solved. That defeats the purpose of the search function.