printing variable length floats

[Ouroborus777]

I've been messing around with printing floats. It seems that printf()
is only capable of printing the fractional portion at a fixed length.
Is there some way to print floats such that the full fraction is shown
but without the trailing zeroes? I realize that this could be an
exercise in string manipulation (ie: stringify the float giving a
length that's longer than the float could possibly be then trim the
trailing zeroes) but I would prefer to do this as modification

What I have ("%.16e"):
-9.0673828125000 000e-001
-4.9169921875000 000e-001
-4.2138671875000 000e-001
-3.6889648437500 000e-001
-2.4206542968750 000e-001
-2.0385742187500 000e-001
0.0000000000000 000e+000
4.3029785156250 000e-003
4.4750976562500 000e-001
7.2753906250000 000e-001
7.8857421875000 000e-001
8.7060546875000 000e-001
8.7890625000000 000e-001
8.9794921875000 000e-001
8.7011718750000 000e-001

What I would like:
-9.0673828125e-1
-4.9169921875e-1
-4.2138671875e-1
-3.68896484375e-1
-2.420654296875e-1
-2.03857421875e-1
0 (or 0.0, or 0.0e+0, but I would prefer just 0)
4.302978515625e-3
4.47509765625e-1
7.275390625e-1
7.8857421875e-1
8.7060546875e-1
8.7890625e-1
8.9794921875e-1
8.701171875e-1

So, any suggestions or maybe point me to some code?

[Ouroborus777]

Re: printing variable length floats

On Aug 23, 5:23 am, Ouroborus777 <deadchic...@gm ail.comwrote:

I've been messing around with printing floats. It seems that printf()
is only capable of printing the fractional portion at a fixed length.
Is there some way to print floats such that the full fraction is shown
but without the trailing zeroes? I realize that this could be an
exercise in string manipulation (ie: stringify the float giving a
length that's longer than the float could possibly be then trim the
trailing zeroes) but I would prefer to do this as modification

err, clicked send to early. I meant to continue:

... but I would prefer to do this as a modification to printf.

[Bartc]

Re: printing variable length floats


"Ouroborus7 77" <deadchicken@gm ail.comwrote in message
news:01ba228a-2c08-46e7-906d-c03f98935008@t1 g2000pra.google groups.com...

I've been messing around with printing floats. It seems that printf()
is only capable of printing the fractional portion at a fixed length.
Is there some way to print floats such that the full fraction is shown
but without the trailing zeroes? I realize that this could be an
exercise in string manipulation (ie: stringify the float giving a
length that's longer than the float could possibly be then trim the
trailing zeroes) but I would prefer to do this as modification
>
What I have ("%.16e"):
-9.0673828125000 000e-001
0.0000000000000 000e+000

What I would like:
-9.0673828125e-1
0 (or 0.0, or 0.0e+0, but I would prefer just 0)

So, any suggestions or maybe point me to some code?

Try using g instead of e.


--
Bartc

[Gordon Burditt]

Re: printing variable length floats

>I've been messing around with printing floats. It seems that printf()

>is only capable of printing the fractional portion at a fixed length.
>Is there some way to print floats such that the full fraction is shown
>but without the trailing zeroes?

The number of digits you get might be surprising. For example, 0.1
as a double ends up with 55 digits, since it can't be represented
exactly. And even the slightest roundoff error can change the
number of digits.

0.1 as long double:
Before: 0.0999999999999 999999945789891 375724778299627 359956502914428 710937500000000 0000000
Value: 0.1000000000000 000000013552527 156068805425093 160010874271392 822265625000000 0000000
After: 0.1000000000000 000000081315162 936412832550558 960065245628356 933593750000000 0000000

0.1 as double:
Before: 0.0999999999999 999916733273153 113259468227624 893188476562500 00
Value: 0.1000000000000 000055511151231 257827021181583 404541015625000 00
After: 0.1000000000000 000194289029309 402394574135541 915893554687500 00

0.1 as float:
Before: 0.0999999940395 355224609375000 000000000000000 000000000000000 00
Value: 0.1000000014901 161193847656250 000000000000000 000000000000000 00
After: 0.1000000089406 967163085937500 000000000000000 000000000000000 00

>I realize that this could be an
>exercise in string manipulation (ie: stringify the float giving a
>length that's longer than the float could possibly be then trim the
>trailing zeroes) but I would prefer to do this as modification
>
>What I have ("%.16e"):
>-9.0673828125000 000e-001
>-4.9169921875000 000e-001
>-4.2138671875000 000e-001
>-3.6889648437500 000e-001
>-2.4206542968750 000e-001
>-2.0385742187500 000e-001
>0.000000000000 0000e+000
>4.302978515625 0000e-003
>4.475097656250 0000e-001
>7.275390625000 0000e-001
>7.885742187500 0000e-001
>8.706054687500 0000e-001
>8.789062500000 0000e-001
>8.979492187500 0000e-001
>8.701171875000 0000e-001
>
>What I would like:
>-9.0673828125e-1
>-4.9169921875e-1
>-4.2138671875e-1
>-3.68896484375e-1
>-2.420654296875e-1
>-2.03857421875e-1
>0 (or 0.0, or 0.0e+0, but I would prefer just 0)
>4.302978515625 e-3
>4.4750976562 5e-1
>7.275390625e-1
>7.8857421875 e-1
>8.7060546875 e-1
>8.7890625e-1
>8.9794921875 e-1
>8.701171875e-1
>
>So, any suggestions or maybe point me to some code?

[Ouroborus777]

Re: printing variable length floats

On Aug 23, 6:02 am, "Bartc" <b...@freeuk.co mwrote:

Try using g instead of e.

Ah, that does the trick. By itself, it still clips to 7 significant
digits but with "%.99g", it works great.

[Joe Wright]

Re: printing variable length floats

Gordon Burditt wrote:

The number of digits you get might be surprising. For example, 0.1
as a double ends up with 55 digits, since it can't be represented
exactly. And even the slightest roundoff error can change the
number of digits.
>
0.1 as long double:
Before: 0.0999999999999 999999945789891 375724778299627 359956502914428 710937500000000 0000000
Value: 0.1000000000000 000000013552527 156068805425093 160010874271392 822265625000000 0000000
After: 0.1000000000000 000000081315162 936412832550558 960065245628356 933593750000000 0000000
>
0.1 as double:
Before: 0.0999999999999 999916733273153 113259468227624 893188476562500 00
Value: 0.1000000000000 000055511151231 257827021181583 404541015625000 00
After: 0.1000000000000 000194289029309 402394574135541 915893554687500 00
>
0.1 as float:
Before: 0.0999999940395 355224609375000 000000000000000 000000000000000 00
Value: 0.1000000014901 161193847656250 000000000000000 000000000000000 00
After: 0.1000000089406 967163085937500 000000000000000 000000000000000 00
>

Your representations are deceptively wide. The 64-bit mantissa of long
double has about 19 digits of precision. The 53-bit double about 16 digits.

--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

[Gordon Burditt]

Re: printing variable length floats

>The number of digits you get might be surprising. For example, 0.1

>as a double ends up with 55 digits, since it can't be represented
>exactly. And even the slightest roundoff error can change the
>number of digits.
>>
>0.1 as long double:
>Before:

>0.099999999999 999999994578989 137572477829962 735995650291442 871093750000000 00000000

>Value:

>0.100000000000 000000001355252 715606880542509 316001087427139 282226562500000 00000000

>After:

>0.100000000000 000000008131516 293641283255055 896006524562835 693359375000000 00000000

>>
>0.1 as double:
>Before: 0.0999999999999 999916733273153 113259468227624 893188476562500 00
>Value: 0.1000000000000 000055511151231 257827021181583 404541015625000 00
>After: 0.1000000000000 000194289029309 402394574135541 915893554687500 00
>>
>0.1 as float:
>Before: 0.0999999940395 355224609375000 000000000000000 000000000000000 00
>Value: 0.1000000014901 161193847656250 000000000000000 000000000000000 00
>After: 0.1000000089406 967163085937500 000000000000000 000000000000000 00
>>

>
>Your representations are deceptively wide. The 64-bit mantissa of long
>double has about 19 digits of precision. The 53-bit double about 16 digits.

I'm still printing the exact value that is present in the floating-point
variable. There's nothing wrong with that, especially when you're
talking about and/or debugging issues of floating-point roundoff.

Granted, whatever calculation produced this value does not have
this much precision.

[Ouroborus777]

Re: printing variable length floats

On Aug 24, 10:18 pm, gor...@hammy.bu rditt.org (Gordon Burditt) wrote:

Granted, whatever calculation produced this value does not have
this much precision.

For the curious, they're from a 3D model file. The values are stored
as 16-bit reals. I'm using POV-Ray to render the data. While I
probably don't need what relatively little precision real16 offers, it
makes me happy in an OCD kind of way.