Rounding

knapak

Hiya I have a variable that shoud return a value of say... 50. Instead it's returning 49.9999999956 which may sound pretty good but I need the 50. I found this thing in the Std library which is an enum called float_round_style that includes something called round_toward_infinity... but I can't find how to actually use the durn code, an example or anything other than the description :mad:. Can anyone tell me how to use it? Is there any other easier way to acomplish what I want? Thanks!!! :cool:

Christian Graus

double d = 49.9999999956; int i = (int)d; Christian Graus - Microsoft MVP - C++

knapak

Nope, that leaves me with 49, not 50.

Shog9 0

ceil()

You must be careful in the forest Broken glass and rusty nails If you're to bring back something for us I have bullets for sale...

Christian Graus

I find that very difficult to comprehend. Numbers should round up from .5 upwards, thought. That makes the floor function totally useless. This will work. #include double d = 49.99999956; int i = (int)ceil(d); Christian Graus - Microsoft MVP - C++

knapak

Thank you! that sure works!!! :-D

knapak

Thank you, that was easy!!! :-D

G_S

The Function "double ceil(double);" will always round UP. that is even if you have 49.00005 the ceil() function will round UP to 50; if you want to round to the nearest Integer you need somthing else. G_S

Christian Graus

G_S wrote: The Function "double ceil(double);" will always round UP. I guess that's why it's called ceil ( as in ceiling ). G_S wrote: if you want to round to the nearest Integer you need somthing else. I was helping someone who wants to round UP. However, casting to int seems to just drop the floating point part, when I assumed it would round to the nearest int, up or down. HEnce, I suggested ceil. Christian Graus - Microsoft MVP - C++

PJ Arends

double d = 49.99999999; int i = (int)(d + 0.5); i will now be properly rounded.

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G_S

PJ Arends wrote: double d = 49.99999999; int i = (int)(d + 0.5); Nice very nice but what about the (-) negative ones test it you'll see. int i; if(d<0) i = (int)(d - 0.5); else i = (int)(d + 0.5); G_S

Shog9 0

To alter the processor's default rounding behavior, you can use the _controlfp() function to set specific FPU flags (such as _RC_CHOP, _RC_UP, _RC_DOWN, _RC_NEAR).

You must be careful in the forest Broken glass and rusty nails If you're to bring back something for us I have bullets for sale...

cmk

The method PJ showed, and you expanded for negative numbers, is a standard way. Shog hinted at another way doing it using _controlfp (or _control87). The rounding control flags specify how the assembly instruction frndint will round. From Intel info on frndint: The Control Word 16-bit register is used ... The RC field (bits 11 and 10) or Rounding Control determines how the FPU will round results in one of four ways: 00 = Round to nearest, or to even if equidistant (this is the initialized state) 01 = Round down (toward -infinity) 10 = Round up (toward +infinity) 11 = Truncate (toward 0) So, by default, it rounds the way you want. The problem is i don't know of any C function that calls this, hence the workaround of adding 0.5 and then truncating by cast to int. A more general (non-portable) solution would be to write your own function that calls frndint. e.g.

__forceinline double FkRound( double N )
{
__asm {
fld N
frndint
}
}

Before calling this you could call _controlfp() to set the way you want rounding to work, or just to make sure it hasn't been changed by some other code. This works for both positive and negative numbers. ...cmk Save the whales - collect the whole set

Bob Stanneveld

Hello, The standard implies that when a floating point value is converted to a intergral value, the fractional part is discarded... I also got the blogging virus..[^]