Calculate by memory value

[chu121su12]

I'm doing my calculator and it's helping me until now. But in some cases, the rounding is not working fine.
Ex: when the result should be 4, the printout becomes 3.9999999 .
I know to avoid this problem is to calculate by the value in memory, instead of it's real value.

My questions:
1. How to access the memory value (read and store a value)?
2. How to do the calculation? (addition, subtract, divide, multiply, power)
3. How to display the result?

Edit: Currently, I'm using double for my value container.

Many questions may follow.

[jsmith]

I don't understand what you believe is the solution. The problem you are encountering is floating point roundoff error. Your best solution, if possible, is to perform the math using integers.

[chu121su12]

Actually, I want to modify my code to work that way. Currently, I'm just using normal operation such as 2 + 3 . I'd want to access the binary value of 2 and 3 so I can operate it as binary (maybe this should be the keyword) before returning as real number.

I may convert these integers to binary, but the exponential numbers are quite difficult...

By the way, how do you calculate double in integer representation (beside of using fraction) ?

[xabnu]

If you wanted to see how floating point numbers are set up, here you go:

http://en.wikipedia.org/wiki/Floating_point
or just google floating point.

If you want a really simple solution, and don't mind limited accuracy (but hey, when will you not have limited accuracy, or infinite significant figures anyway?), you can tell your printing command to round to the nearest so many digits.

[chu121su12]

Thank you. But actually, I want to learn about this. So, it's not about accuracy anymore.

This is an extension for my q1.

Let say a double variable contains number 2.0. How can I access the binary representation for this number?

If the 2.0 is 0000 0010 (is it?) in computer memory, how can I get the value so I can compute it?

[chu121su12]

Well, I just realize, you guys are answering my question for the rounding, not for the programming part. I'm sorry if my point was not clear before.

So, back to my questions above.

[exception]

You can get bit access to double values like described here:
http://www.cplusplus.com/forum/articles/3827/

If the 2.0 is 0000 0010 (is it?) in computer memory, how can I get the value so I can compute it?

It's the binary representation of the integer 2; the IEEE 756 double value is stored differently; see the above link.

And about that accuracy: use a multi-precision rational type. Speed won't be an issue for a calculator. If you require exact root calculations, there is a data type LEDA::real which can handle that (it's not free, though. And if you want to write it yourself, you will require some solid mathematical knowledge about algebraic numbers and their construction (there is a beautiful constructive proof that the algebraic numbers are a field, using the Sylvester resultant, which would help you))

[chu121su12]

Hmm... I remember that it require pointers to access the bits.
For the calculation, I may be able to solve it -somehow-. But now, I just need to work on obtaining the bits.

I'll continue my work then and I'll come back after some stressess...
If you guys have any suggestion, please post it.

So for now, thank you for your replies.

[xabnu]

The way I usually access a single bit is bitwise logic or shifting. Given the standard for floating point numbers, you would need to know the significance of the bit(s) you are testing, i.e. sign exponent mantissa.

Using integers, it's a bit more obvious. If (a & b) is not zero, then the numbers both have at least one bit in common that is 1.

int a,b;
b = 2 * 2;
printf("What is a's 3rd bit? %s", (a & b)? "It's a 1\n" : "It's a 0\n");

[chu121su12]

@xabnu
As I thought, integers are easier to handle. Thanks for the reply.

After some thinking, I'm now planning to use strings and fractions. These bits playing are quite hard for me.

One thing - My code: vector_calculatorXX.cpp @
http://www.mediafire.com/?sharekey=5e10097ada3bbd65ab1eab3e9fa335ca4881245c386baa3c
Don't forget to download conio1.2.h. Or use conio2.h.

[chu121su12]

Found it! If this is correct, so it is ^^

i = *(int*)&value;

[chu121su12]

I found a simple equation:

0.01234-0.01334 gives you -0.000999999999999999 in my calculator, while the correct answer should be -0.001

cool, isn't it?