As easy as pi?

1. Use [code][ /code] tags!
2. float data type is accurate up to 6 digits, while double is up to 15. That's exactly what you see in your output.

Cheers

Hi,

I was using the following program to learn about the unary scope resolution operator.

However the output of the problem was as follows:

Local float value of PI = 3.1415927410125732
Global double value of PI = 3.14159265358979

I was very surprised that the float value and hte double values were so different can anyone explain why this is so?

Kind Regards

Baz

//Fig. 3.24: fig03_24.cpp
//Using the unary scope resolution operator
#include <iostream>

using std::cout;
using std::endl;

#include <iomanip>

using std::setprecision;

//define global constant PI
const double PI = 3.14159265358979;

int main()
{
//define global contant PI
const float PI = static_cast< float >(::PI);

//display values of local and global PI constants
cout << setprecision (20)
<< " Local float value of PI = " << PI
<< "\nGlobal double value of PI = " << ::PI << endl;

return 0; //indicates successful termination
} //end main

Hi thanks ashukasama,

I thought it must be something to do with casting from a higher order varaible to a lower order one.

However from your response:

float
4 bytes, IEEE 754. Covers a range from 1.40129846432481707e-45 to 3.40282346638528860e+38 (positive or negative).

surely float still covers up to 45 decimal places which should be more than enough to cover the 16 decimal places used in the original double variable.

I just thought that float should be accurate enough in this instance?

Or have I missed the point here??

Baz

[ Merged and moved threads ]

@to Bazman
Please don't post this kind of problems in Visual C++ forum!
Please don't open multiple threads for the same problem!
Please use CODE tags (http://www.dev-archive.com/forum/misc.php?do=bbcode#code)!

Thank you!
Ovidiu

Hi,

I apologise that I posted this twice.

But I had no replies in my mailbox and I couldn't even find the thread when I looked last night so I thought somehow I hadn't posted it correctly.

Also I will endeavour to post to correct forum it future. Which forum should this have gone in?

In any case thanks to all of you its clear now.

Baz

ah OK apologies again!

will teach me to read the whole list next time!

To be sure to represent exactly any binary FP number with a decimal representation, the same number of significand decimal digits than significand binary digits is required, in the worst case.
This is due to the fact that 10==2*5.
2 appears only once in the prime number development of 10.

Notation: x**n is x at the power of n.
Proof:
1/(2**n) is represented 0.0...01 in binary, where the number of zeroes is equal to n.
It's easy to demonstrate, with a recurrency argument, that the decimal representation of this number contains n significand digits and is terminated by 5.

1/(2**1) -> 0.5
1/(2**2) -> 0.25
1/(2**3) -> 0.125

1/(2**n) -> 0.0...0A...B5
1/(2**(n+1)) -> 0.0...0C...D25

32 bits float numbers TYPICALLY have a binary representation with a 24 bits mantissa, a 1 bit sign, and a 7 bits signed exponent.
Their precision is 24 binary digits which is the same precision as 24*log10(2)==7.224... decimal digits.

But, to represent exactly this FP number, sometimes, 24 significant decimal digits would be necessary!

Typically, double is 64 bits, with 53 bits for the mantissa, 1 bit for the sign and 10 bits for the exponent.