Hexadecimal to decimal conversion problem.Also, how to convert a char number to an actual int number

Question

Please help me to identify the error in this program, as for me it's looking correct,I have checked it,but it is giving wrong answers. In this program I have checked explicitly for A,B,C,D,E,F,and according to them their respective values.

[Edited]:Also,this question relates to how a character number is converted to actual integer number.

#include<iostream>
#include<cmath>
#include<bits/stdc++.h>
using namespace std;
void convert(string num)
{
   long int last_digit;
    int s=num.length();
    int i;
    long long int result=0;
    reverse(num.begin(),num.end());                 
    for(i=0;i<s;i++)
    {
        if(num[i]=='a' || num[i]=='A')
        {
            last_digit=10;
            result+=last_digit*pow(16,i);
        }
        else if(num[i]=='b'|| num[i]=='B')
        {
            last_digit=11;
            result+=last_digit*pow(16,i);
        }
        else if(num[i]=='c' || num[i]=='C')
        {
            last_digit=12;
            result+=last_digit*pow(16,i);
        }
        else if(num[i]=='d'|| num[i]=='D' )
        {
            last_digit=13;
            result+=last_digit*pow(16,i);
        }
        else if(num[i]=='e'|| num[i]=='E' )
        {
            last_digit=14;
            result+=last_digit*pow(16,i);
        }
        else if(num[i]=='f' || num[i]=='F')
        {
            last_digit=15;
            result+=last_digit*pow(16,i);
        }
        else {
            last_digit=num[i];
        result+=last_digit*pow(16,i);
        }
    }
    cout<<result;
}
int main()
{
    string hexa;
    cout<<"Enter the hexadecimal number:";
    getline(cin,hexa);
    convert(hexa);
}

Answer 1

Your code is very convoluted and wrong.

You probably want this:

void int convert(string num)
{
  long int last_digit;
  int s = num.length();
  int i;
  long long int result = 0;

  for (i = 0; i < s; i++)
  {
    result <<= 4;                     // multiply by 16, using pow is overkill
    auto digit = toupper(num[i]);     // convert to upper case

    if (digit >= 'A' && digit <= 'F')
      last_digit = digit - 'A' + 10;   // digit is in range 'A'..'F'
    else
      last_digit = digit - '0';        // digit is (hopefully) in range '0'..'9'

    result += last_digit;
  }

  cout << result;
}

But this is still not very good:

• the function should return a long long int instead of printing the result
• a few other thing can be done mor elegantly

So a better version would be this:

#include <iostream>
#include <string>

using namespace std;

long long int convert(const string & num)  // always pass objects as const & if possible
{
  long long int result = 0;

  for (const auto & ch : num)      // use range based for loops whenever possible
  {
    result <<= 4;
    auto digit = toupper(ch);

    long int last_digit;           // declare local variables in the inner most scope

    if (digit >= 'A' && digit <= 'F')
      last_digit = digit - 'A' + 10;
    else
      last_digit = digit - '0';

    result += last_digit;
  }

  return result;
}

int main()
{
  string hexa;
  cout << "Enter the hexadecimal number:";
  getline(cin, hexa);
  cout << convert(hexa);
}

There is still room for more improvements as the code above assumes that the string to convert contains only hexadecimal characters. Ideally a check for invalid characters should be done somehow. I leave this as an exercise.

The line last_digit = digit - 'A' + 10; assumes that the codes for letters A to F are contiguous, which in theory might not be the case. But the probability that you'll ever encounter an encoding scheme where this is not the case is close to zero though. The vast majority of computer systems in use today use the ASCII encoding scheme, some use EBCDIC, but in both of these encoding schemes the character codes for letters A to F are contiguous. I'm not aware of any other encoding scheme in use today.

Answer 2

Your problem is in the elsecase in which you convert num[i] from char to its ascii equivalent. Thus, for instance, if you try to convert A0, the 0is converted into 48 but not 0. To correct, you should instead convert your num[i] into its equivalent integer (not in asci).

To do so, replace :

else {
            last_digit=num[i];
        result+=last_digit*pow(16,i);

with

  else {
            last_digit = num[i]-'0';
            result+=last_digit*pow(16,i);
    }

In the new line, last_digit = num[i]-'0'; is equivalent to last_digit = (int)num[i]-(int)'0';which substracts the representation code of any one-digit-number from num[i] from the representation code of '0'

It works because the C++ standard guarantee that the number representation of the 10 decimal digits are contiguous and in incresing order (official ref iso-cpp and is stated in chapter 2.3 and paragraph 3

Thus, if you take the representation (for instance the ascii code) of any one-digit-number num[i] and substract it with the representation code of '0' (which is 48 in ascii), you obtain directly the number itself as an integer value.

An example of execution after the correction would give:

A0
160

F5
245

A small codereview: You are repeating yourself with many result+=last_digit*pow(16,i);. you may do it only once at the end of the loop. But that's another matter.

Answer 3

You are complicating the problem more than you need to (std::pow is also kinda slow). std::stoul can take a numerical base and automatically convert to an integer for you:

#include <string>
#include <iostream>

std::size_t char_count{0u};
std::string hexa{};
std::getline(std::cin, hexa);
hexa = "0x" + hexa;
unsigned long value_uint = std::stoul(hexa, &char_count, 16);