Binary to decimal conversion not working

[SuddeN]

Im having trouble getting this function to work. I need to take an 8-bit binary number in as a string and convert it to decimal. I already have a function to validate the binary number, making sure it's all 1's or 0's, and only 8 bits long. I can't get the conversion to work though. If I enter "11111111" as my string it should return 255 but it returns 128. When I go through debugging, it shows that my if statement resolves to false at biNum.substr positions 0-6, only resolving to true at the last position. Any help would be amazing. I've been working on this for 3 days and can't get anywhere.

  int ConvertFromBinary(string biNum, int len)
{
	int num = 0;

	for (int i = 0; i < len; i++)
	{
		if (biNum.substr(i) == "1")
		{
			num = num + pow(2, i);
		}
		
	}

	return num;
}

[lastchance]

Change
if (biNum.substr(i) == "1")
to
if (biNum[i] == '1')
(Note the single quotes as well as the simple index.) substr doesn't do what you appear to think it does.

Other things:
- you don't need to pass len as a parameter; biNum is a std::string so just use biNum.size()
- pow( 2, i ) actually returns a double; you will probably get away with it, but this sort of thing is a recipe for floating-point truncation errors
- in future, please give us complete code so that we can test it without having to write in all the headers and a driver routine.

[SuddeN]

That worked perfectly! Thank you so much!

[hbcpp]

you should go and check out what substr does.

However, with only that change made, still the conversion wouldn't work correctly

I have changed the function and it seems to work fine, check it out

  int ConvertFromBinary(string biNum)
{
	int num = 0;

	int len = biNum.size();

	int real_pos = len - 1;

	for (int i = 0; i < len; i++)
	{
		if (biNum[i] == '1')
		{
			num = num + pow(2, real_pos);
		}

		--real_pos;
	}

	return num;
}

[SuddeN]

I noticed that after implementing the first change. Thank you though!

[jonnin]

some little stuff...
pow is kinda slow for integer powers. many of us keep a custom "ipow" for integers for speed, and to avoid any of the mentioned roundoff issues. Here, a lookup table might do.
num+= …
is the same as
num = num+...
and you should use this shorthand. logic and arithmetic operators have a version eg -+ *= ^= and so on

its possible to write it without the if, such that it just adds zero when zero comes in and adds 2^n when a 1 comes in. This is good for practice, to see if you can do it, as sometimes this is a useful trick. Here, its just the numeric value ('1'-'0' or '0'-'0') * the pow expression, or in Boolean, (str[x]=='1')*pow...

[lastchance]

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

int ConvertFromBinary( string biNum )
{
   int num = 0;
   int len = biNum.size();
   for ( int i = 0; i < len; i++ )
   {
      if ( biNum[i] == '1' ) num += 1 << len - 1 - i;
   }
   return num;
}

int main()
{
   string tests[] = { "11111111", "1100", "0011", "101010", "10111" };
   for ( string s : tests ) cout << s << "  " << ConvertFromBinary( s ) << '\n';
}

You could also cheat and use a bitset.

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

unsigned long ConvertFromBinary( string biNum )
{
   bitset<32> B( biNum );
   return B.to_ulong();
}

int main()
{
   string tests[] = { "11111111", "1100", "0011", "101010", "10111" };
   for ( string s : tests ) cout << s << "  " << ConvertFromBinary( s ) << '\n';
}