C++ Convert a number from base A to base B

This is a C++ class which has a member function that takes 3 arguments and converts the number from first base to second base. It takes the following arguments.

– std::string thestlstring

– int base

– int base2

The number is stored in the form of std::string and returns the number as string . It supports bases 2 till base 16.

Source code is contributed by: massiveattack92(at)hotmail.com

#include <cmath>
#include <string>
#include <sstream>

#include "inputs.h"

inputs::inputs(std::string newstring, int newbase, int newbase2) {
  inpthenumber = newstring;
  inpbase = newbase;
  inpbase2 = newbase2;
}

inputs::~inputs(void) {}

std::string inputs::conv(std::string thestlstring, int base, int base2) {

  // variable declarations

  int myval;
  int remainder;
  int loopvar = 0;
  int mynumber = 0;
  int testsum = 0;
  float value1;
  float value2;
  char mychar;
  char element;
  std::string stlstring;
  std::stringstream thestream;

  // generate base 10 integer from number in base 'mybase'

  string::iterator pos;
  pos = thestlstring.end();

  for (--pos; pos >= thestlstring.begin(); --pos) {
    mychar = * pos;

    switch (mychar) {
    case '0':
      myval = 0;
      break;
    case '1':
      myval = 1;
      break;
    case '2':
      myval = 2;
      break;
    case '3':
      myval = 3;
      break;
    case '4':
      myval = 4;
      break;
    case '5':
      myval = 5;
      break;
    case '6':
      myval = 6;
      break;
    case '7':
      myval = 7;
      break;
    case '8':
      myval = 8;
      break;
    case '9':
      myval = 9;
      break;
    case 'A':
      myval = 10;
      break;
    case 'B':
      myval = 11;
      break;
    case 'C':
      myval = 12;
      break;
    case 'D':
      myval = 13;
      break;
    case 'E':
      myval = 14;
      break;
    case 'F':
      myval = 15;
      break;
    }
    mynumber += myval * (int) std::pow(base, loopvar);

    ++loopvar;
  }
  // generate 'n' - (n-1) = # of digits in base 10 number
  // generated above
  for (int n = 0; testsum < mynumber; ++n) {
    testsum = testsum + (int) std::pow(base2, n) * (base2 - 1);
  }
  // declare array for insertion of digits from the
  // converted number
  char * revarray = new char[n - 1];

  // generate the new number
  for (int i = 0; i < n; ++i) {
    value1 = (float)((float) mynumber / (float) base2);
    value2 = (float) floor((float) mynumber / (float) base2);
    remainder = mynumber - (int) value2 * base2;

    if (remainder < 10) element = char(remainder + '0');
    else if (remainder == 10) element = 'A';
    else if (remainder == 11) element = 'B';
    else if (remainder == 12) element = 'C';
    else if (remainder == 13) element = 'D';
    else if (remainder == 14) element = 'E';
    else if (remainder == 15) element = 'F';
    // insert an element each time within the loop 
    revarray[i] = element;
    // move down one base multiple 
    mynumber = (mynumber - remainder) / base2;
  }
  // create std::string format of converted number 
  for (int i2 = n - 1; i2 >= 0; --i2) {
    thestream << revarray[i2];
  }
  thestream >> stlstring;
  // return the number
  return stlstring;
}

//----------------------
//inputs.h
//----------------------

#include <string>

class inputs {
  private:

    std::string inpthenumber;
  int inpbase;
  int inpbase2;

  public:
    inputs(std::string, int, int); // called
  // automatically when instance created

  ~inputs(void); // automatically called
  // when object released from RAM
  // must return void.

  std::string conv(std::string thestlstring, int base, int base2);

};

Sample Output

Input: thestlstring = “777”, base = 8, base2 = 10
Output: 511
Explanation: 777 in octal (base =8) when converted to decimal (base =10) is 511.

Input: thestlstring = “927”, base = 10, base2 = 16
Output: 39F
Explanation: 927 in decimal (base =10) when converted to hexadecimal (base = 16) is 39F.

You can verify these values using a scientific calculator program available in windows OS.