Compare/Convert Float to Text

David Crow

You need to compare the difference to some epsilon. Check float.h for DBL_EPSILON or FLT_EPSILON.

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mohit 12

thx all for your support.. nvm it i solved it ..stupid ME ... just added this method: bool classname::isStringEquals(float fNewNumber,float fOldNumber) { CString value1; CString value2; value1.Format( _T("%.4f"), fNewNumber); value2.Format( _T("%.4f"), fOldNumber); if(value1==value2) return true; else return false; cheers from canada

Maximilien

bad answer, bad solution. You are rounding off values and killing performance. Read this : What Every Computer Scientist Should Know About Floating-Point Arithmetic[^] Max.

Watched code never compiles.

Iain Clarke Warrior Programmer

You're kinda doing it wrong - at least in general. You code may work fine for your requirements! Try:

BOOL IsEqualish (double d1, double d2, double dEpsilon)
{
ASSERT(dEpsilon > 0.0);
d1 -= d2;
if (d1 < 0.0)
d1 = -d1; // or just d1 = abs (d1);
return (d1 <= dEpsilon)
}

or for consise:

BOOL IsEqualish (double d1, double d2, double dEpsilon) { return abs (d1 - d2) <= dEpsilon; }

Where dEpsilon is the error margin. For 4 dps, dEpsilon = 0.0001, and so on. Iain.

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Aescleal

I'm not a great one for going mental about code's performance but that's just horrible on so many levels. Why not do something like:

bool near_enough_equals( double a, double b )
{
return (b - a) < ( (b + a) / 2000.0 );
}

Cheers, Ash

Niklas L

Why use strings? Just round the values.

double a = 1.00003;
double b = 0.99997;
a *= 100;
b *= 100;
a += 0.5;
b += 0.5;

if (::floor(a) == ::floor(b))
{
// Wooha!
}

home

Luc Pattyn

using strings for numeric information when you don't need strings is never the right approach. For numerical problems, use a numeric solution. Here are some possibilities: 1. you might put an upper limit on the deviation, similar to:

float is=0.999999877543;
float closeTo=1.0;
bool closeEnough=abs(is-closeTo)<0.01;

that would be close enough within 1% (since closeTo itself is one). 2. you might opt for a relative deviation (assuming the ideal value isn't zero):

float is=12.999999877543;
float closeTo=13.0;
bool closeEnough=abs((is-closeTo)/closeTo)<1.0e-6;

that would be close enough when within 1 ppm (pars pro million) :)

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mohit 12

How am i killing performance, well all i wanted to check is if the other float value is close or 0.10 and that code does check that i cannot use greater or less than since both float can be changed might be >0 or <0 , so rounding off values check if it's close to 0.10 or not because i do not have problem in my math if value if 0.0999978 or 0.10 both will give me same answer since it just checks value and then do not use them for any calculation, i will paste my code later on here and then you can have a look at it and you might be able to give me better suggestion, thx for reply tho cheers from canada.

mohit 12

I didn't get your code why would i use epsilon ? and it has to be FLOAT thx for reply. cheers from canada

mohit 12

I am dealing with float and there is not way i can use DOUBLE for that particular code and my value might be negative so it might return negative value using your code and also b will be always less than a so negative :doh: thx for reply cheers from canada

mohit 12

floor will convert 0.099999787 to 0.10 ? and it has to be float value will it work ? thx cheers from canada.

mohit 12

as i mentioned in previous post those value can be changed so and it may vary from negative to zero and zero to +2.707 E06 so not sure if that would work.

Luc Pattyn

of course it can work. Whatever you can achieve with strings can be done without strings. FYI: electronic circuit simulators use a combination of absolute and relative tolerance, so they would do a convergence test similar to the following:

float previous=...;
float current=...;
float threshold=0.1e-3; // an absolute tolerance of 0.1 millivolt for small voltages
if (abs(previous)>1) threshold+=abs(previous)*100.0e-6; // a relative tolerance of 100 ppm
bool converged = abs(current-previous) < threshold;

in reality the two magic constants are variables, as the user is allowed to choose their value. :)

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Tim Craig

mohit`12 wrote:

if(value1==value2) return true; else return false;

Ok, setting aside your real problem, why not simply code this as: return (value1 == value2); Call me anal if you like, but that's so much simpler and easier to read.

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Iain Clarke Warrior Programmer

mohit`12 wrote:

I didn't get your code why would i use epsilon ?

You yourself brought up the problem of floating point numbers being not perfect - if you read some of the links given you, you will realise that testing for equality is basically a waste of time. Digital computers simply do not do perfect representations of non-integers. Unless you're very lucky, and it's 1/(2^n). Two numbers can be reeeeeeeally close, and not equal. The simple fact you asked your original question shows that you know this already.

mohit`12 wrote:

and it has to be FLOAT

So change double to float. Or whatever FLOATs your boat. Not the hardest of challenges! Iain. ps, My code will also cope (as it should) with negative floating values.

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Niklas L

a = 0.09999;
a *= 100; // a = 9.999
a += 0.5; // a = 10.499
floor(a); // a = 10.0

b = 0.10001;
b *= 100; // b = 10.001
b += 0.5; // b = 10.051
floor(b); // b = 10.0

home

Aescleal

Floats are converted to doubles before any serious arithmetic is done on them. The only reason you'd use a float is if you really had to save the 4 bytes (on most implementations). You actually introduce more arithmetic errors by saving everything as a float between calculations than doing everything as a double. As for the negative problem, use std::fabs or something on the values before processing them:

bool are_these_two_near_enough( double a, double b )
{
a = std::fabs( a ); b = fabs( b );

return std::fabs( a - b ) < ( ( a + b ) / 2000.0 );

}

and if you're really worried about the difference between float and double use this as the function signature:

template< typename T >
bool are_these_two_floating_point_values_near_enough( T a, T b )

Edited as I didn't have all the HTML escapes properly sorted

modified on Tuesday, August 24, 2010 5:54 AM

Lost User

... and use "ceil" to round up value for example,

printf("ceil of 2.3 is %.1lf\n", ceil (2.3));

output

ceil of 2.3 is 3.0

to use the floor/ceil should be included math.h Sorry for my poor english. Best Regards, Alexander S.

Niklas L

?

home

Lost User

float a = 1.555;
float b = floor(a); // b = 1
float c = ceil(a); // c = 2

a = 0.09999;
b = floor(a); // b = 0
c = ceil(a); // c = 1

modified on Tuesday, August 24, 2010 1:56 AM