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Answer 1

A:

Can you multiply by 100 and then truncate to an integer? Then you could format the result like one would with dollars and cents. Simply dividing by 100 might land you back at square one due to floating-point representation issues.

Kristo 2010-03-31 20:31:58

Not really. Standard double floating point representations are accurate to 15 digits. So unless the OP is dealing with numbers in the ten billions of dollars range he should be fine. In any case, multiplication runs into the same floating-point representation issues.

Billy ONeal 2010-03-31 20:43:07

Answer 2

+2 A:

sprintf it into a buffer, and then put the NUL char two bytes past the '.'

Then printf your final string using the intermediate one.

quixoto 2010-03-31 20:32:27

Answer 3

A:

What about using double trunc(double) from GLibC?

Romain 2010-03-31 20:34:16

Because not everybody is using GLibC.

Billy ONeal 2010-03-31 20:35:41

@Billy The question doesn't say "I don't want to use GLibC", so answer is valid, right ? :)

Romain 2010-03-31 20:46:58

Yes, it's valid. That's why I didn't downvote. That doesn't necessarily make it good. :)

Billy ONeal 2010-03-31 20:48:53

@Billy: I never attempted to argue it was a *good* solution ;)

Romain 2010-04-03 20:58:53

Answer 4

+10 A:

#include <math.h>
...
    printf("%.2f", floor(100 * R) / 100);

Hans Passant 2010-03-31 20:35:48

+1 but since the question is tagged C++ possibly also provide the C++ iostream way?

Mark B 2010-03-31 20:53:50

Are there some odd cases where this gives the wrong answer? I'm not certain, but does there exist a floating point value f and an integer n such that f is less than the exact value of n/100, but the floating point result of f*100 is greater than or equal to n? Of course you can question whether it matters that f is truncated "incorrectly", since in "base 10 space" it represents a range which straddles a 0.01 marker.

Steve Jessop 2010-03-31 21:06:53

Are negative numbers possible? With -99.999, multiply to -9999.9, floor is -10000, and the result is -100.

David Thornley 2010-03-31 21:30:10

@Steve: yes, kind of: it does not work with 90071992547410.016 (which is printed as 90071992547410.02).

RaphaelSP 2010-03-31 21:40:27

@David: I guess it should be `(R > 0 ? floor(100 * R) : ceil(100 * R)) / 100`

RaphaelSP 2010-03-31 21:41:42

Answer 5

+4 A:

All you have to do is subtract .005 from the number and magically printf will behave as you wish: always round down.

frankc 2010-03-31 20:42:41

Unless the number is negative. Then it will blow up.

Billy ONeal 2010-03-31 20:43:49

You beat me to the edit...it doesn't blow up. It just does the reverse. In this case, add .005.

frankc 2010-03-31 20:45:14

Doesn't that seem a bit more complicated than nobugz' answer for no gain? Sure, you save a division and multiplication, but you trade it for a branch dependence (which is worse), and code that much less clearly expresses your intent.

Billy ONeal 2010-03-31 20:47:18

So use `R + (R>0 ? -0.005 : 0.005)`, basically...

Romain 2010-03-31 20:48:45

@BillyONeal: what makes you so sure that the performance difference between a division and a branch is (a) definitely one way or the other, (b) significant when immediately followed by a call to `printf`? I'd have thought it would at least depend on whether the platform has floating-point hardware or not: division is likely not going to be cheaper than a single branch if it's done in software.

Steve Jessop 2010-03-31 20:58:32

@Steve Jessop: Perhaps that's platform dependent, but on the machines most of us are using, things are being done in hardware. But in any case, the intent is more important here. user275455's solution is non-obvious, nobugz' is obvious. That alone is compelling reason to use that method.

Billy ONeal 2010-03-31 21:05:05

@Billy maybe, maybe not. I don't it's any less clear than any other solution. All of them are likely to include the comment /* force printf to truncate to two decimal places */ so it seems irrelevant. If your domain is definitely restricted to either positive or negative numbers, then I think the solution is clearly this best performing.

frankc 2010-03-31 21:09:05

Not to mention, the multiplication solution is more likely to suffer from overflow.

frankc 2010-03-31 21:12:01

@user275455: it won't overflow in any situation where it hasn't already lost precision long, long ago. It's meaningless to round a number to a multiple of .01 if it's within a factor of 100 of FLOAT_MAX.

Steve Jessop 2010-03-31 21:15:53

Answer 6

A:

Another solution, using casts:

...
printf("%.2lf", (double) ((int) (R * 100)) / 100);

Juliano 2010-03-31 21:10:46

This isn't well-defined if R can be negative.

David Thornley 2010-03-31 21:30:58

I've tested in Borland C++ Builder and it worked for both negative and positive values.

Juliano 2010-03-31 21:52:14

Answer 7

A:

Another way, truly sign agnostic: printf("%d.%d\n", (int) r ,abs((int)(r*100) % 100));

frankc 2010-03-31 21:38:09

Answer 8

+2 A:

If you have it, use fmod() to chop the tail of the double:

double rounded = R - fmod(R, 0.01);
// Now just print rounded with what you were using before

This has the advantage of working the same if R is positive or negative.

Donal Fellows 2010-03-31 21:55:04

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Truncating double without rounding in C

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