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# Floating point calculation

I'm working in an ARM (ARM9) system which does not have Floating point
co-processor or Floating point libraries. But it does support long long int
(64 bits).
Can you provide some link that would discuss about ways to emulate floating
point calculations with just long int or long long int. For eg., if i've a
formula X=(1-b)*Y + b*Z in floating point domain, i can calculate X with
just long ints (but, some data may be lost in final division; That's OK)

Floating Point:
X=(1-b)*Y + b*Z
/* 'b' is a floating point variable with 4 points precision and 'b' is in
the range of 0 to 1;X, Y and Z are unsigned int*/

With long int:
I can emulate the above calculation as:
X=((10000-10000*b)*Y +10000*b*Z)/10000

I'm in need of some link that would discuss this and any similar approach.

--
-Vinoth

Nov 14 '05 #1
7 3352

Vinoth wrote:
I'm working in an ARM (ARM9) system which does not have Floating point co-processor or Floating point libraries. But it does support long long int (64 bits).
Can you provide some link that would discuss about ways to emulate floating point calculations with just long int or long long int. For eg., if i've a formula X=(1-b)*Y + b*Z in floating point domain, i can calculate X with just long ints (but, some data may be lost in final division; That's OK)
Floating Point:
X=(1-b)*Y + b*Z
/* 'b' is a floating point variable with 4 points precision and 'b' is in the range of 0 to 1;X, Y and Z are unsigned int*/

With long int:
I can emulate the above calculation as:
X=((10000-10000*b)*Y +10000*b*Z)/10000

I'm in need of some link that would discuss this and any similar approach.
--
-Vinoth

One way of faking floating point that I thought of is to find free
multi-precision math libraries -- GNU mp and the library that comes
with GNU bc and dc come to mind -- since those libraries treat the
numbers as arrays of digits.

Gregory Pietsch

Nov 14 '05 #2
Vinoth wrote:
I'm working in an ARM (ARM9) system which does not have Floating point
co-processor or Floating point libraries. But it does support long long int
(64 bits).
Can you provide some link that would discuss about ways to emulate floating
point calculations with just long int or long long int. For eg., if i've a
formula X=(1-b)*Y + b*Z in floating point domain, i can calculate X with
just long ints (but, some data may be lost in final division; That's OK)

Floating Point:
X=(1-b)*Y + b*Z
/* 'b' is a floating point variable with 4 points precision and 'b' is in
the range of 0 to 1;X, Y and Z are unsigned int*/

With long int:
I can emulate the above calculation as:
X=((10000-10000*b)*Y +10000*b*Z)/10000

I'm in need of some link that would discuss this and any similar approach.

Hi, I'm a bit loth to step in here (reading this lurking in c.l.c, not a
C expert), but couldn't you implement floating-point using those long
longs? Write fmul, fdiv, fadd etc functions that mask off a long long
into sign, exponent and mantissa and deal with them.

Multiplication is like in standard index form (multiply the mantissa,
add the exponent) and with adding you multiply so the numbers have the

There is an IEEE specification for floating point (e.g. google IEEE
floating-point) that includes rules for what the bit patterns mean,
representation of small numbers (a special case for numbers between -1
and 1), infinities etc as well - probably better than coming up with
your own scheme. I don't know if this is worth the effort for you, or
if there are drawbacks I've not thought of, but I don't see why you
couldn't do all this in standard C.

http://stevehollasch.com/cgindex/coding/ieeefloat.html

HTH, all the best,
Rob M

--
Rob Morris: arr emm four four five [at] cam dot ac dot uk
Nov 14 '05 #3

Vinoth wrote:
I'm working in an ARM (ARM9) system which does not have Floating point
co-processor or Floating point libraries. But it does support long long int
(64 bits).
Can you provide some link that would discuss about ways to emulate floating
point calculations with just long int or long long int. For eg., if i've a
formula X=(1-b)*Y + b*Z in floating point domain, i can calculate X with
just long ints (but, some data may be lost in final division; That's OK)

Floating Point:
X=(1-b)*Y + b*Z
/* 'b' is a floating point variable with 4 points precision and 'b' is in
the range of 0 to 1;X, Y and Z are unsigned int*/

With long int:
I can emulate the above calculation as:
X=((10000-10000*b)*Y +10000*b*Z)/10000

I'm in need of some link that would discuss this and any similar approach.

Your "emulation" should work fine, if the products and
sum in the numerator don't grow too large for `long'. If
you know enough about the ranges of Y and Z to be sure this
won't happen, all is well. If not, you can use `long long'
for the intermediate results:

X = ((10000LL - 10000LL*b) * Y + 10000LL*b * Z) / 10000LL;

There are a number of possible improvements you may want
to consider. The first is to get rid of those `10000LL*b'
computations, which is easy: instead of storing `b' itself,
store `10000 * b' in a `long' variable called `B':

X = ((10000LL - B) * Y + (long long)B * Z) / 10000LL;

Rearranging the expression with a little algebra can
eliminate one of the multiplications and permit a little more
of the computation to use plain `long' instead of `long long'
(which may be faster, especially if `long long' is emulated
in software):

X = Y + (long)((Z - Y) * (long long)B / 10000LL);

If you change the scaling factor from 10000 to something
that's a power of two, you can replace the division with a
shift. 16384 (1 << 14) is pretty close to your original
10000, so assuming that `B' is now `b * 16384' you'd have

X = Y + (long)( ((Z - Y) * (long long)B) >> 14 );

There's a potential trap here: if `Z - Y' is negative
so the product being shifted is also negative, C doesn't
specify exactly what happens with the right shift. Since
you're only concerned with one implementation you could
check whether it does what you want. If it doesn't, or
if you want to be sure the code will work elsewhere, too,
you could make sure that no negative numerators appear:

if (Z >= Y)
X = Y + (long)( ((Z - Y) * (long long)B) >> 14 );
else
X = Y - (long)( ((Y - Z) * (long long)B) >> 14 );

This is about as far as you can go with portable C --
which is a shame, really, because some machines are capable
of better. For example, there may be an instruction (or
instruction sequence) to multiply two 32-bit numbers and
yield a 64-bit product, but C cannot multiply two `long's to
get a `long long'. If you used 32-bit scaling instead of
the 14 bits shown above, the second term would simply be the
high-order 32 bits of the 64-bit product and the machine might
be able to extract it without shifting, but C has no portable
way to perform such dissections. It's possible that a smart
optimizing compiler might be able to exploit such capabilities
of the machine (I'd especially recommend looking into the
possibility of 32-bit scaling), but there are no guarantees.

What you're doing with the "emulation" is called "fixed-
point arithmetic," and the techniques can be applied in more
sophisticated form -- to get a properly-rounded answer, for
example, or to deal with numbers that have both integer and
fractional parts. A small amount of research may give you
some good ideas ...

--
Er*********@sun.com

Nov 14 '05 #4
Thanks to all for the information. I'm intrested in trying out all basic
operations on fixed-point arithmetic. Can you point to some free library

"Eric Sosman" <er*********@sun.com> wrote in message
news:d6**********@news1brm.Central.Sun.COM...

Vinoth wrote:
I'm working in an ARM (ARM9) system which does not have Floating point
co-processor or Floating point libraries. But it does support long long
int
(64 bits).
Can you provide some link that would discuss about ways to emulate
floating
point calculations with just long int or long long int. For eg., if i've
a
formula X=(1-b)*Y + b*Z in floating point domain, i can calculate X with
just long ints (but, some data may be lost in final division; That's OK)

Floating Point:
X=(1-b)*Y + b*Z
/* 'b' is a floating point variable with 4 points precision and 'b' is in
the range of 0 to 1;X, Y and Z are unsigned int*/

With long int:
I can emulate the above calculation as:
X=((10000-10000*b)*Y +10000*b*Z)/10000

I'm in need of some link that would discuss this and any similar
approach.

Your "emulation" should work fine, if the products and
sum in the numerator don't grow too large for `long'. If
you know enough about the ranges of Y and Z to be sure this
won't happen, all is well. If not, you can use `long long'
for the intermediate results:

X = ((10000LL - 10000LL*b) * Y + 10000LL*b * Z) / 10000LL;

There are a number of possible improvements you may want
to consider. The first is to get rid of those `10000LL*b'
computations, which is easy: instead of storing `b' itself,
store `10000 * b' in a `long' variable called `B':

X = ((10000LL - B) * Y + (long long)B * Z) / 10000LL;

Rearranging the expression with a little algebra can
eliminate one of the multiplications and permit a little more
of the computation to use plain `long' instead of `long long'
(which may be faster, especially if `long long' is emulated
in software):

X = Y + (long)((Z - Y) * (long long)B / 10000LL);

If you change the scaling factor from 10000 to something
that's a power of two, you can replace the division with a
shift. 16384 (1 << 14) is pretty close to your original
10000, so assuming that `B' is now `b * 16384' you'd have

X = Y + (long)( ((Z - Y) * (long long)B) >> 14 );

There's a potential trap here: if `Z - Y' is negative
so the product being shifted is also negative, C doesn't
specify exactly what happens with the right shift. Since
you're only concerned with one implementation you could
check whether it does what you want. If it doesn't, or
if you want to be sure the code will work elsewhere, too,
you could make sure that no negative numerators appear:

if (Z >= Y)
X = Y + (long)( ((Z - Y) * (long long)B) >> 14 );
else
X = Y - (long)( ((Y - Z) * (long long)B) >> 14 );

This is about as far as you can go with portable C --
which is a shame, really, because some machines are capable
of better. For example, there may be an instruction (or
instruction sequence) to multiply two 32-bit numbers and
yield a 64-bit product, but C cannot multiply two `long's to
get a `long long'. If you used 32-bit scaling instead of
the 14 bits shown above, the second term would simply be the
high-order 32 bits of the 64-bit product and the machine might
be able to extract it without shifting, but C has no portable
way to perform such dissections. It's possible that a smart
optimizing compiler might be able to exploit such capabilities
of the machine (I'd especially recommend looking into the
possibility of 32-bit scaling), but there are no guarantees.

What you're doing with the "emulation" is called "fixed-
point arithmetic," and the techniques can be applied in more
sophisticated form -- to get a properly-rounded answer, for
example, or to deal with numbers that have both integer and
fractional parts. A small amount of research may give you
some good ideas ...

--
Er*********@sun.com

Nov 14 '05 #5

Vinoth wrote:
Thanks to all for the information. I'm intrested in trying out all basic
operations on fixed-point arithmetic. Can you point to some free library

Wow! You must be an awfully fast reader to have
studied those "about 26,200" results in less than three
hours! I'm afraid I can't offer more help than those
26,200 articles can -- and since you've already found
Sorry.

(You might also want to Google for "top-posting."
The "about 84,400" articles won't take *you* very long,
and may convey something useful.)

--
Er*********@sun.com

Nov 14 '05 #6
On 19 May 2005 "Vinoth" <no*********@emailaddress.com> wrote:
Thanks to all for the information. I'm intrested in trying out all basic
operations on fixed-point arithmetic. Can you point to some free library

[Previous messages removed]

Please do not top post to newsgroups.

---druck

--
The ARM Club Free Software - http://www.armclub.org.uk/free/
The 32bit Conversions Page - http://www.quantumsoft.co.uk/druck/
Nov 14 '05 #7
Vinoth wrote:

Thanks to all for the information. I'm intrested in trying out all basic
operations on fixed-point arithmetic. Can you point to some free library

http://www.arm.com/documentation/App...tes/index.html

contains some a basic introduction on implementing fixed point
binary arithmetic on ARM cores.

Chris
Nov 14 '05 #8

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