float variables in loops

[vijay]

Hello,

What happens to float variable in loops. For example,

float f=8.7;
if(f<8.7)
printf("less");
else if(f==8.7)
printf("equal");
else if(f>8.7)
printf("more");

prints "less". Shouldn't it print "equal".

regards,
vijay.

 

[Christian Bau]

Hello,

What happens to float variable in loops. For example,

float f=8.7;
if(f<8.7)
printf("less");
else if(f==8.7)
printf("equal");
else if(f>8.7)
printf("more");

prints "less". Shouldn't it print "equal".

No, it shouldn't.

What is the type of f?
What is the type of 8.7?
Are they the same types?

 

[Jens.Toerring]

What happens to float variable in loops. For example,

float f=8.7;
if(f<8.7)
printf("less");
else if(f==8.7)
printf("equal");
else if(f>8.7)
printf("more");

prints "less". Shouldn't it print "equal".

No, you can't expect that. The basic problem is that numbers are
stored with a finite number of digits. And numbers like 8.7 are
no simple looking numbers anymore when converted to binary but
have an infinite number of digits (like one third is an infinite
fraction when written in base 10). But since you only have a
finite numbers of bits to store them in, the numbers get truncated.
So most floating point numbers can be stored only as an approxi-
mation and what's stored of your 8.7 is probably something like
8.6999998 instead of 8.7 - you can easily see that effect when
you try to print it out with enough digits, try e.g.

printf( "%20.18f\n", f );

Moreover, since (usually) floats have less bits than doubles,
the same number stored in a float and a double variable will
differ - just try it out with

printf( "%20.18f %20.18f\n", f, 8.7 );

This also indicates that "8.7" isn't treated as a float but
as a double - all calculations in C are done per default in
double and all constants like "8.7" are treated as doubles -
to avoid that you would have to write "8.7f" instead. So what
you do in your program is comparing the value of the float
variable 'f' with the value of 8.7 when stored as a double.
But since these values typically differ you hardly ever will
get "less" printed out (except for numbers that can be stored
using a small number of digits, your result would be different
if you would use e.g. 2.5 instead of 8.7).

So the first thing to keep in mind when dealing with floating
point numbers is that comparing them for equaliy will typically
only work by accident.
Regards, Jens

 

[Krishanu Debnath]

Hello,

What happens to float variable in loops. For example,

float f=8.7;
if(f<8.7)
printf("less");
else if(f==8.7)
printf("equal");
else if(f>8.7)
printf("more");

prints "less". Shouldn't it print "equal".

regards,
vijay.

FAQs already address this issue.

Please read followings ..

http://www.eskimo.com/~scs/C-faq/q14.4.html
http://www.eskimo.com/~scs/C-faq/q14.5.html


Krishanu

 

[vijay]

No, you can't expect that. The basic problem is that numbers are
stored with a finite number of digits. And numbers like 8.7 are
no simple looking numbers anymore when converted to binary but
have an infinite number of digits (like one third is an infinite
fraction when written in base 10). But since you only have a
finite numbers of bits to store them in, the numbers get truncated.
So most floating point numbers can be stored only as an approxi-
mation and what's stored of your 8.7 is probably something like
8.6999998 instead of 8.7 - you can easily see that effect when
you try to print it out with enough digits, try e.g.

printf( "%20.18f\n", f );

Moreover, since (usually) floats have less bits than doubles,
the same number stored in a float and a double variable will
differ - just try it out with

printf( "%20.18f %20.18f\n", f, 8.7 );

8.699999809265136719
8.699999999999999289

This is the output I get. How is it really calculated? Does the output
depend upon the compiler(i'm using gcc), system etc.

printf( "%20.18f %20.18f\n", f, 8.7f );
gives the same output, maybe because now it treats 8.7 as a float and
not as a double.

regards,
vijay.

 

[Tim Prince]

8.699999809265136719
8.699999999999999289

This is the output I get. How is it really calculated? Does the output
depend upon the compiler(i'm using gcc), system etc.

If your system uses open source libraries, such as glibc or newlib, you have
the opportunity to look up how printf() et al. do their work. Or, look up
ideas in a fine reference, such as Plauger's "Standard C Library." On many
systems, where there is a choice of gcc or other compilers, the same
printf() would be shared by multiple compilers. A library will comply with
IEEE standard if it gets 17 digits right, for the standard 64-bit double.
Accuracy of additional digits may depend on whether your system supports
long double.

 

[Lawrence Kirby]

On Wed, 04 May 2005 08:11:51 +0000, Jens.Toerring wrote:

....

printf( "%20.18f %20.18f\n", f, 8.7 );

This also indicates that "8.7" isn't treated as a float but
as a double - all calculations in C are done per default in
double

That was true for K&R C but is not true in standard C where operations
involving operands of type float or float in combination with
integer operands produce a float result.

The default argument promotions, where (amongst other things) a float
argument gets promoted to double in an unprototyped function call or in a
variable argument list, still exist in standard C but don't affect
"calculations".

Lawrence

 

[Chris Croughton]

8.699999809265136719
8.699999999999999289

This is the output I get. How is it really calculated? Does the output
depend upon the compiler(i'm using gcc), system etc.

It depends on the respresentation of float and double on your system,
which will be related to the compiler and the hardware. The standard
says only that a float must have at least 6 digits of precision and a
double at least 10 digits, but they could be identical on some systems
as long as they meet the minimum criteria.

printf( "%20.18f %20.18f\n", f, 8.7f );
gives the same output, maybe because now it treats 8.7 as a float and
not as a double.

I assume you mean that both outputs are the same. Yes, if you just say
8.7 it is interpreted as a double (6.4.4.1 para 4), if you say 8.7f then
it will be a float.

In general, however, comparing floating point numbers for exact equality
is not sensible, because rounding can occur without you knowing it. For
instance, 9.7 - 8.7 == 1.0 may well fail, as may (x+1) - 1 == x. In
general the way to test is for "near equality", choose some value
EPSILON and do something like:

if (abs(expression) < EPSILON)
/* near enough */

or even write it as a function so that it auto-scales epsilon according
to numbers being compared:

#include <float.h>
#include <math.h>

int float_compare(float a, float b)
{
float epsilon = (fabs(a) + fabs(b)) * FLT_EPSILON;
if (a - b > epsilon) return +1;
if (b - a > epsilon) return -1;
return 0;
}

Chris C

 

[Kevin Bracey]

In message <[email protected]>

On Wed, 04 May 2005 08:11:51 +0000, Jens.Toerring wrote:

...


That was true for K&R C but is not true in standard C where operations
involving operands of type float or float in combination with
integer operands produce a float result.

That's a half-truth. Yes, the result has semantic type float, but it's
allowed to have extra precision. Thus in:

float f;
double d = f * f;

the final result d may have full double precision, or it may just have
float precision, depending on FLT_EVAL_METHOD. So the traditional
mode of operation where "all calculations in C are done per default in
double" is still an implementation option.

And, even more bizarrely - something I didn't know until recently - even
constants might have more precision than their type. So

double d = 8.7F;

could leave you with full double precision. Bleurgh. The only way to
guarantee a single-precision constant is:

float f = 8.7F;
or
double d = (float) 8.7F;