Hi,
Why does Math.Sqrt() only accept a double as a parameter? I would think
it would be just as happy with a decimal (or int, or float, or ....). I can
easily convert back and forth, but I am interested in what is going on behind
the scenes and if there is some aspect of decimals that keep them from being
used in this calculation.
Thanks!
Ethan
Ethan Strauss Ph.D.
Bioinformatics Scientist
Promega Corporation
2800 Woods Hollow Rd.
Madison, WI 53711
608-274-4330
800-356-9526 et***********@p romega.com 13  10504 
Ethan Strauss <Et**********@d iscussions.micr osoft.comwrote:
Why does Math.Sqrt() only accept a double as a parameter? I would think
it would be just as happy with a decimal (or int, or float, or ....). I can
easily convert back and forth, but I am interested in what is going on behind
the scenes and if there is some aspect of decimals that keep them from being
used in this calculation.
Well, decimals are typically used when you're dealing with numbers
which are naturally and *exactly* represented as decimals - currency
being the most common example.
The kind of number you're likely to take the square root of is the kind
of number you should probably be using double for instead.
--
Jon Skeet - <sk***@pobox.co m>
Web site: http://www.pobox.com/~skeet
Blog: http://www.msmvps.com/jon.skeet
C# in Depth: http://csharpindepth.com
Ethan Strauss wrote:
Why does Math.Sqrt() only accept a double as a parameter? I would think
it would be just as happy with a decimal (or int, or float, or ....). I can
easily convert back and forth, but I am interested in what is going on behind
the scenes and if there is some aspect of decimals that keep them from being
used in this calculation.
Usually that type of functions returns a value of the same
type as itself.
Sqrt of a decimal will not return a true decimal (decimal is
supposed to be exact).
Sqrt of an int will definitely not return an int.
double Sqrt(decimal) and double Sqrt(int) will not follow
practice for functions
decimal Sqrt(decimal) and int Sqrt(int) will give an
impression about the result that is not correct
Besides: when do you need to take the square root of 87 dollars ??
:-)
Arne
All of the variables used by the Math class are primitive types.
The CLR does not consider a Decimal to be a primitive type (the CLR does not
contain special IL instructions to handle decimal types). I would imagine
that is one of the reason…. maybe???
You can check out the definition of a Decimal and you will see that it
implements/overrides just about all the math operators on it including but
not limited to +,-,*,/,Round, Ceiling, Floor etc, many of this are also
included on the Math class but the Decimal type provides its own
implementation.
"Ethan Strauss" <Et**********@d iscussions.micr osoft.comwrote in message
news:77******** *************** ***********@mic rosoft.com...
Hi,
Why does Math.Sqrt() only accept a double as a parameter? I would think
it would be just as happy with a decimal (or int, or float, or ....). I
can
easily convert back and forth, but I am interested in what is going on
behind
the scenes and if there is some aspect of decimals that keep them from
being
used in this calculation.
Thanks!
Ethan
Ethan Strauss Ph.D.
Bioinformatics Scientist
Promega Corporation
2800 Woods Hollow Rd.
Madison, WI 53711
608-274-4330
800-356-9526
et***********@p romega.com
Ethan,
Why did you think they ever created a double (or value representations like
that with other names).
(As in the beginning all data was stored only binary in bytes and a while
later as so called binary decimals).
In my idea it was to do things like Math.Sqrt()
Cor
"Ethan Strauss" <Et**********@d iscussions.micr osoft.comschree f in bericht
news:77******** *************** ***********@mic rosoft.com...
Hi,
Why does Math.Sqrt() only accept a double as a parameter? I would think
it would be just as happy with a decimal (or int, or float, or ....). I
can
easily convert back and forth, but I am interested in what is going on
behind
the scenes and if there is some aspect of decimals that keep them from
being
used in this calculation.
Thanks!
Ethan
Ethan Strauss Ph.D.
Bioinformatics Scientist
Promega Corporation
2800 Woods Hollow Rd.
Madison, WI 53711
608-274-4330
800-356-9526
et***********@p romega.com
On Jun 5, 1:44 am, Arne Vajhøj <a...@vajhoej.d kwrote:
Ethan Strauss wrote:
Why does Math.Sqrt() only accept a double as a parameter? I would think
it would be just as happy with a decimal (or int, or float, or ....). I can
easily convert back and forth, but I am interested in what is going on behind
the scenes and if there is some aspect of decimals that keep them from being
used in this calculation.
Usually that type of functions returns a value of the same
type as itself.
Sqrt of a decimal will not return a true decimal (decimal is
supposed to be exact).
Decimal operations are exact/accurate in certain well-defined
circumstances.
There are plenty of operations on decimal which won't give exact
results:
1m / 3 for example. Likewise even addition - both 1e25 and 1e-25 are
exactly
representable as decimals, but their sum isn't.
Sqrt of an int will definitely not return an int.
Indeed, but then you wouldn't really want to.
double Sqrt(decimal) and double Sqrt(int) will not follow
practice for functions
decimal Sqrt(decimal) and
double Sqrt(int)
would be fine in my view, but the first is useless for typical
encouraged uses of decimal, and the second is already available
through an implicit conversion from int to double.
decimal Sqrt(decimal) and int Sqrt(int) will give an
impression about the result that is not correct
Does decimal division give you that impression as well? How about
integer division?
Besides: when do you need to take the square root of 87 dollars ??
And *that* is the real reason, IMO. The encouraged uses of decimal are
for the kinds of quantity one just doesn't take square roots of (like
money, as per your example).
Jon
On Jun 5, 4:08 am, "Rene" <a...@b.comwrot e:
All of the variables used by the Math class are primitive types.
Round, Floor, Ceiling, Max, Min, Truncate, Sign and Abs all have
overloads which take decimals.
The CLR does not consider a Decimal to be a primitive type (the CLR does not
contain special IL instructions to handle decimal types). I would imagine
that is one of the reason…. maybe???
Well, doing a square root properly (as opposed to converting to
double, taking the square root and then converting back, which would
be a horrible way to go) would certainly be rather slower than when
using double due to the lack of hardware support. More importantly
though, it just wouldn't be useful for the intended uses of decimal.
Jon
Round, Floor, Ceiling, Max, Min, Truncate, Sign and Abs all have
overloads which take decimals.
Yes, but if you look at the implementation of these overloads, they are
nothing more that a wrapper to the Decimal class. For example: Math.Round
will internally call Decimal.Round
>Well, doing a square root properly (as opposed to converting to
double, taking the square root and then converting back, which would
be a horrible way to go) would certainly be rather slower than when
using double due to the lack of hardware support. More importantly
though, it just wouldn't be useful for the intended uses of decimal.
Exactly, my point was that since there are no special IL instructions for
Decimals, getting the square root of a decimal using decimal context wouldn’t
be very efficient.
Rene <a@b.comwrote :
Round, Floor, Ceiling, Max, Min, Truncate, Sign and Abs all have
overloads which take decimals.
Yes, but if you look at the implementation of these overloads, they are
nothing more that a wrapper to the Decimal class. For example: Math.Round
will internally call Decimal.Round
Sure, but that doesn't change my point at all. You claimed that all the
methods in Math took primitive types (a parameter is a variable, after
all), and the methods above are counterexamples .
The implementation should be irrelevant to the discussion - it could
just as easily have been the other way round, with Decimal.Round
calling Math.Round.
Well, doing a square root properly (as opposed to converting to
double, taking the square root and then converting back, which would
be a horrible way to go) would certainly be rather slower than when
using double due to the lack of hardware support. More importantly
though, it just wouldn't be useful for the intended uses of decimal.
Exactly, my point was that since there are no special IL instructions for
Decimals, getting the square root of a decimal using decimal context wouldn=3Ft
be very efficient.
And *my* point was that efficiency isn't the main issue here. Just
because it wouldn't be efficient to take the square root doesn't make
it undesirable per se. It's the uses of decimal which make it
undesirable.
--
Jon Skeet - <sk***@pobox.co m>
Web site: http://www.pobox.com/~skeet
Blog: http://www.msmvps.com/jon.skeet
C# in Depth: http://csharpindepth.com
Jon Skeet [C# MVP] wrote:
On Jun 5, 1:44 am, Arne Vajhøj <a...@vajhoej.d kwrote:
>Ethan Strauss wrote:
>> Why does Math.Sqrt() only accept a double as a parameter? I would think
it would be just as happy with a decimal (or int, or float, or ....). I can
easily convert back and forth, but I am interested in what is going on behind
the scenes and if there is some aspect of decimals that keep them from being
used in this calculation.
Usually that type of functions returns a value of the same
type as itself.
Sqrt of a decimal will not return a true decimal (decimal is
supposed to be exact).
Decimal operations are exact/accurate in certain well-defined
circumstances.
There are plenty of operations on decimal which won't give exact
results:
1m / 3 for example.
True, but division would be missed if it was not there.
I would not have a problem if decimal division gave an exception
if the result was not exact, but that would probably be too
inefficient to test for that.
Likewise even addition - both 1e25 and 1e-25 are
exactly
representable as decimals, but their sum isn't.
Which is not good either.
But I see your point.
Decimal is not exact in other contexts either.
I love the concept of decimal, but I hate the implementation chosen.
>Sqrt of an int will definitely not return an int.
Indeed, but then you wouldn't really want to.
>double Sqrt(decimal) and double Sqrt(int) will not follow
practice for functions
decimal Sqrt(decimal) and
double Sqrt(int)
would be fine in my view, but the first is useless for typical
encouraged uses of decimal, and the second is already available
through an implicit conversion from int to double.
>decimal Sqrt(decimal) and int Sqrt(int) will give an
impression about the result that is not correct
Does decimal division give you that impression as well?
It can.
How about
integer division?
No, integer division works just as expected. I prefer the Pascal style
with one operator for floating point division and another for integer
division to emphasize that it is two different operators.
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