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Mar 19th, 2005

Adding decimal numbers together?

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Hi ladies and gents,

I'm trying the next exercise:

Write a function value, wich adds the decimal numbers of 'n' together and return the result of this into main? For example:

0.1234 = 10
0.01234 = 10
0.3286 = 19

What Ive got so far is this:

C Syntax (Toggle Plain Text)
short amount (float n)
{
	short a = 0, i, totvalue = 0;
 
	for (i = 0; i < sizeof (n); i++)
	{
		a = n = (n * 10);
		n -= a;
		totvalue += a;
	}
 
	return totvalue;
}
 
int main()
{
	double a = 0.1234;
 
	cout<<"Value = " << amount(a) <<endl;
 
	return 0;
}
short amount (float n)
{
	short a = 0, i, totvalue = 0;

	for (i = 0; i < sizeof (n); i++)
	{
		a = n = (n * 10);
		n -= a;
		totvalue += a;
	}

	return totvalue;
}

int main()
{
	double a = 0.1234;

	cout<<"Value = " << amount(a) <<endl;

	return 0;
}

Problem is, when I run threw the loop in the function and arrive at the fourth digit '4', the compiler changes the number into '39997' and I get a wrong result :o

Also, is it possible to change the working of the loop using binary operators?

Thanks for the help :!:

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Adding binary numbers in C++

JoBe

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Mar 19th, 2005

Re: Adding decimal numbers together?

I would convert the number to a string using sprintf() then loop through each character in the string and check with isdigit(). If true, then atoi() the character and add it to the sum. An interesting exercise!

vegaseat

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Mar 19th, 2005

Re: Adding decimal numbers together?

Thanks for the reply Vegaseat, but untill this part of exercises, there is no mentioning of using your examples in the piece of the book that Ive done yet, I'll have to go about 60 pages further before the explanation of strings is mentioned in detail so that I could try what you are suggesting

I'm at part 4 (classes and functions) :lol:
Strings is in part 6 ( Functions, strings en arrays)

JoBe

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Mar 19th, 2005

Re: Adding decimal numbers together?

Quote originally posted by JoBe ...

Problem is, when I run threw the loop in the function and arrive at the fourth digit '4', the compiler changes the number into '39997' and I get a wrong result :o

Floating point behaves this way because the value stored is an approximation. Since you're passing a double, you may want to receive it as a double rather than a float. Then you may discover...

Using sizeof (n) isn't the way to go about it either. That is the bytes required for the underlying storage, not the number of digits beyond the decimal place.

Dave Sinkula

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Mar 20th, 2005

Re: Adding decimal numbers together?

Hi Dave,

Changing the variable from float into double doesn't change anything, so, your hint towards:

Quote ...

Then you may discover...

isn' found?

I understand what your saying about sizeof(), problem is, I don't have a clue as to how to change this towards the amount of numbers behind the 0,... ( don't know the name of ',' in English)

I think (almost certain) that I have to find a way to determine how many numbers are behind the ',' problem is, I don't know how I can do that

JoBe

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Mar 20th, 2005

Re: Adding decimal numbers together?

>don't know the name of ',' in English
Radix

>I don't know how I can do that
Floating-point is a bitch to work with. You would be better off converting to a string as vegaseat suggested:

C Syntax (Toggle Plain Text)
#include <iostream>
#include <sstream>
#include <string>
 
using namespace std;
 
int precision (double val)
{
  int n = 0;
 
  ostringstream out;
  out<< val;
  string s = out.str();
 
  return s.length() == 1 ? 0 : s.length() - 2;
}
 
int main()
{
  cout<< precision ( 0.0 ) <<endl;
  cout<< precision ( 0.1 ) <<endl;
  cout<< precision ( 0.12 ) <<endl;
  cout<< precision ( 0.123 ) <<endl;
  cout<< precision ( 0.1234 ) <<endl;
  cout<< precision ( 0.12345 ) <<endl;
  cout<< precision ( 0.123456 ) <<endl;
}
#include <iostream>
#include <sstream>
#include <string>

using namespace std;

int precision (double val)
{
  int n = 0;

  ostringstream out;
  out<< val;
  string s = out.str();

  return s.length() == 1 ? 0 : s.length() - 2;
}

int main()
{
  cout<< precision ( 0.0 ) <<endl;
  cout<< precision ( 0.1 ) <<endl;
  cout<< precision ( 0.12 ) <<endl;
  cout<< precision ( 0.123 ) <<endl;
  cout<< precision ( 0.1234 ) <<endl;
  cout<< precision ( 0.12345 ) <<endl;
  cout<< precision ( 0.123456 ) <<endl;
}

Narue

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Mar 20th, 2005

Re: Adding decimal numbers together?

Hi Narue,

Thanks for the help, tough, why is it that in your example, the decimal numbers aren't adding up,

Quote ...

cout<< precision ( 0.123456 ) <<endl;

This for example shows as result 6 and not 21?

Also, why do you use int n = 0, it seems you don't use it in the function?

Quote ...

return s.length() == 1 ? 0 : s.length() - 2;

I understand this to be a selection, but I don't understand what it's doing?

I added this piece of code into the program that I wrote

Quote ...

if ( n >= 1){ a = n; n -= a; a = 0;}

This is because previously when I entered a number like 1,1234 it didn't get rid of the number before the radix.
Not that it matters since it seems I'm not going to solve this exercise with the code I wrote :-|

I see you, Dave and Vegaseat became moderators, very smart from Dani to do this, you guys are a real asset towards this C++ community, congrats to you all :!:

JoBe

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Mar 20th, 2005

Re: Adding decimal numbers together?

>why is it that in your example, the decimal numbers aren't adding up
Because it's your project and not mine. I was just showing you the simplest way of determining how many digits of precision there are.

>why do you use int n = 0, it seems you don't use it in the function?
Leftovers from an older version.

>I understand this to be a selection, but I don't understand what it's doing?
If there are no digits past the radix, subtracting 2 from the string length would give you -1 instead of the desired 0. The equivalent is:

C Syntax (Toggle Plain Text)
if ( s.length() == 1 )
  return 0;
else
  return s.length() - 2;
if ( s.length() == 1 )
  return 0;
else
  return s.length() - 2;

Narue

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Mar 21st, 2005

Re: Adding decimal numbers together?

>Because it's your project and not mine. I was just showing you the simplest way of determining how many digits of precision there are.
Got it, thanks for the help Narue :!:

JoBe

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Mar 23rd, 2005

Re: Adding decimal numbers together?

Hi Narue,

Ive been trying to get this exercise to work with the piece of code you gave me, but I'm stuck :!:

Ive put numbers 1, 2 & 3 left towards the code that I have trouble with

C Syntax (Toggle Plain Text)
#include "stdafx.h"
#include <iostream>
#include <sstream>
#include <string>
 
using namespace std;
 
short amount (double val)
{
	short a = 0, i, totvalue = 0;
	ostringstream out;
 
	out<< val;
	string s = out.str();
 
	for (i = 0; i < s.length(); ++i) // Number 1
	{			
		a = s * 10; // Number 2
		s = s - a; // Number 3
		totvalue += a;
	}
 
	return totvalue;
}
 
int main()
{
	double a = 0.999;
 
	cout<<"Value = " << amount(a) <<endl;
 
	return 0;
}
#include "stdafx.h"
#include <iostream>
#include <sstream>
#include <string>

using namespace std;

short amount (double val)
{
	short a = 0, i, totvalue = 0;
	ostringstream out;

	out<< val;
	string s = out.str();

	for (i = 0; i < s.length(); ++i) // Number 1
	{			
		a = s * 10; // Number 2
		s = s - a; // Number 3
		totvalue += a;
	}

	return totvalue;
}

int main()
{
	double a = 0.999;

	cout<<"Value = " << amount(a) <<endl;

	return 0;
}

1) I believe I can use this to determine the amount of digits behind the radix, correct?

2 & 3) when using this code, I get the following error C2676: binary '*' or '-': 'class std::basic_string<char,struct std::char_traits<char>,class std::allocator<char> >' does not define this operator

I understand it's not recognizing this operator, but, how can I get it to recognize it?

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