Binary to Decimal W/ the presentation of equation

[xoox]

Im trying to make a program that converts binary to decimal but having the needs to show the process of the convertion (Sample image below)

https://ibb.co/jyptyxw

here is my current code
i have no idea how to show the process of the conversion like in the given image.

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
     int bin, dec = 0, remainder, num, base = 1;
     cout << "Enter the binary number: ";
     cin >> num;
     bin = num;
     while (num > 0)
     {
         rememainder = num % 10; //get the last digit of the input
         dec = dec + remainder * base;
         base = base * 2;
         num = num / 10;
     }
     cout << "Decimal Value" << bin << " : " << dec << endl;
     return 0;

}

[Repeater]

Make a blank string for the output.

In your code, the first time you get the last digit, that's 2^0.
In your code, the second time you get the last digit, that's 2^1.
In your code, the third time you get the last digit, that's 2^2.
In your code, the fourth time you get the last digit, that's 2^3.
In your code, the fifth time you get the last digit, that's 2^4.
And so on.

So each time you get the last digit, if it's a 1, put:
1*(2^ whichever) +
on the front of the output string.

When you're done, show the output string.

[xoox]

1*(2^ whichever) +

in this block what variable is 'whichever'
should i use a forloop to counter the exponential increase while checking if its a 0 or a 1 with if condition?

[Repeater]

whichever is the variable you are using to keep track of how many times you've fetched the last digit.

[xoox]

Im actually near to getting it but im stuck with the counter showing up ascending rather than descending.

im having this:
https://ibb.co/XsXvnMp
while it should be like this:
https://ibb.co/jyptyxw

this is my current code :

#include <iostream>
#include <math.h>
using namespace std;

int main()
{
     int bin, dec = 0, remainder, num, base = 1,counter = 0;
     cout << "Enter the binary number: ";
     cin >> num;
     bin = num;
     while (num > 0)
     {
         remainder = num % 10; //get the last digit of the input
         dec = dec + remainder * base;
         base = base * 2;
         num = num / 10;
         counter++;
         if (num % 10 == 1)
         {
         	cout << "1*(2^" << counter << ") ";
		 }
     }
     
     cout << "\nThe decimal equivalent of " << bin << " : " << dec << endl;
     return 0;

}

[Repeater]

Make a blank string for the output.

In your code, the first time you get the last digit, that's 2^0.
In your code, the second time you get the last digit, that's 2^1.
In your code, the third time you get the last digit, that's 2^2.
In your code, the fourth time you get the last digit, that's 2^3.
In your code, the fifth time you get the last digit, that's 2^4.
And so on.

So each time you get the last digit, if it's a 1, put:
1*(2^ whichever) +
on the front of the output string.

When you're done, show the output string.

[xoox]

Im sorry but i dont get it,
i cant store a counter which is an integer inside a string variable for the output
what do you mean in front of the output string?

[Repeater]

i cant store a counter which is an integer inside a string variable for the output

http://www.cplusplus.com/reference/string/to_string/

string output; // Create a new string. Blank

// Each time round the loop, put new letters on the front 
output = string("+ 1*(2^") + to_string(counter) + ") " + output;

// when all done, remove leading '+' from string and output the string 

[xoox]

is there another way to do this without using the to_string function?

[Repeater]

Lots. Pick one.

https://stackoverflow.com/questions/5590381/easiest-way-to-convert-int-to-string-in-c

[jonnin]

you can do a lookup table of all 255 byte combinations without using to-string. Just be endian aware. You can do it off nibbles if you prefer, its just a bit more weird, might use a bit field struct over a byte to do it..

[xoox]

I got it but having trouble fixing the value in the middle part
expected output:
1*(2^4) + 1*(2^2) + 1*(2^1)
what im having:
https://ibb.co/Z8hK34M

this is my current code

#include <iostream>
#include <math.h>
#include <string>

using namespace std;

int main()
{
     int bin, dec = 0, remainder, num, base = 1,counter=0,counter2=0, constnum;
     cout << "Enter the binary number: ";
     cin >> num;
     bin = num;
     constnum = num;
     while(bin > 0)
     {
     	bin=bin/10;
		counter2++;
	 }
	 
     while (num > 0)
     {
     	if (num % 10 == 1) {
		    cout << " 1*(2^" << counter2 << ") +";
		    counter2--;	
			}
		 else if(num % 10 == 0) {
		 	counter2--;
		 }
         remainder = num % 10; //get the last digit of the input
         dec = dec + remainder * base;
         base = base * 2;
         num = num / 10;
         
	}

     cout << "\nThe decimal equivalent of " << constnum << " : " << dec << endl;
     return 0;

}

[Repeater]

Don't output the final '+' on the last time round the loop

[lastchance]

#include <iostream>
#include <string>
using namespace std;

string mathSum( const string &bin )
{
   int n = bin.size();   if ( !n ) return "";
   string rem = bin.substr( 1 );
   return ( bin[0]=='1' ? "2^" + to_string(n-1) + ( rem.find('1') != string::npos ? " + " : "" ) : "" ) + mathSum( rem );
}


int main()
{
   string s;
   s = "1011010";            cout << s << " = " << mathSum( s ) << '\n';
   s = "101101101101101";    cout << s << " = " << mathSum( s ) << '\n';
}

or non-recursive version:

#include <iostream>
#include <string>
using namespace std;

string mathSum( const string &bin )
{
   string result;
   int n = bin.size();
   for ( int i = 0; i < n; i++ )
   {
      if ( bin[n-1-i] == '1' ) result = "2^" + to_string( i ) + ( result != "" ? " + " : "" ) + result;
   }
   return result;
}


int main()
{
   string s;
   s = "1011010";            cout << s << " = " << mathSum( s ) << '\n';
   s = "101101101101101";    cout << s << " = " << mathSum( s ) << '\n';
}

1011010 = 2^6 + 2^4 + 2^3 + 2^1
101101101101101 = 2^14 + 2^12 + 2^11 + 2^9 + 2^8 + 2^6 + 2^5 + 2^3 + 2^2 + 2^0