Division Confusion

[Dave Brown]

Hi all,
I'm having trouble understanding which data type to use to get results I
expect.
using float, decimal, and double I have tried the following, ..

float test = 55/60;

I always get 0 as the results.
I expect 0.916666

I dont understand why its rounding down. Any pointers anybody ?
Rgds,

Dave

[Ed Kaim [MSFT]]

Either the numerator or deniominator (or both) must be floating point values
in order to get a floating point result. For example, 55.0/60 or 55/60.0 or
55.0/60.0.

"Dave Brown" <dave.AT.dbws.net> wrote in message
news:ev**************@TK2MSFTNGP11.phx.gbl...

Hi all,
I'm having trouble understanding which data type to use to get results I
expect.
using float, decimal, and double I have tried the following, ..

float test = 55/60;

I always get 0 as the results.
I expect 0.916666

I dont understand why its rounding down. Any pointers anybody ?
Rgds,

Dave

[Frank Oquendo]

Dave Brown wrote:

float test = 55/60;

I always get 0 as the results.
I expect 0.916666

You're divinding two ints, so you're performing integer division

--
There are 10 kinds of people. Those who understand binary and those who
don't.

http://code.acadx.com
(Pull the pin to reply)

[Dave Brown]

Wow fast replys, thanks guys.

so even though test is a float, it performs integer division, Hmmmph I can
see thats one to really look out for. the value 55 is being passed in as an
int so i just changed it to
float test = (float)minues / 60;

and yes the result is 0.91111.

Thanks again, seems strange behaviour though.

rgds,

Dave
"Frank Oquendo" <fr****@acadxpin.com> wrote in message
news:e7**************@TK2MSFTNGP10.phx.gbl...

Dave Brown wrote:

float test = 55/60;

I always get 0 as the results.
I expect 0.916666

You're divinding two ints, so you're performing integer division

--
There are 10 kinds of people. Those who understand binary and those who
don't.

http://code.acadx.com
(Pull the pin to reply)

[Peter van der Goes]

"Dave Brown" <dave.AT.dbws.net> wrote in message
news:eH*************@TK2MSFTNGP12.phx.gbl...

Wow fast replys, thanks guys.

so even though test is a float, it performs integer division, Hmmmph I can see thats one to really look out for. the value 55 is being passed in as an int so i just changed it to
float test = (float)minues / 60;

and yes the result is 0.91111.

Thanks again, seems strange behaviour though.

rgds,

Dave
"Frank Oquendo" <fr****@acadxpin.com> wrote in message
news:e7**************@TK2MSFTNGP10.phx.gbl...

Dave Brown wrote:

float test = 55/60;

I always get 0 as the results.
I expect 0.916666

You're divinding two ints, so you're performing integer division

--
There are 10 kinds of people. Those who understand binary and those who
don't.

http://code.acadx.com
(Pull the pin to reply)

The determination of the result type is based on the operand types, so no
matter what type of variable you store the result in, it's an integer result
in your example. As Ed points out, all you have to do in the case of
literals is add ".0" to one of them to "promote" it to a double. Then when
performing an arithmetic operation on dissimilar types, the lesser (integer)
is promoted to the greater's (double) type and the result is a double. You
then get the precision you originally expected.

--
Peter [MVP Academic]

[Martin Maat [EBL]]

"Dave Brown" <dave.AT.dbws.net> wrote in message
news:eH*************@TK2MSFTNGP12.phx.gbl...

Wow fast replys, thanks guys. so even though test is a float, it performs integer division, Hmmmph I can see thats one to really look out for. the value 55 is being passed in as an int so i just changed it to
float test = (float)minues / 60;

and yes the result is 0.91111.

Thanks again, seems strange behaviour though.

Not at all. Look at it this way: first the expression 55 / 60 is evaluated.
At that time the compiler doesn't know or care what you want to use the
result for. It is not that smart. So the only relevant part to the first
operation is

55 / 60

The compiler does have a preference for Int32 types so whenever it can it
will treat constants as Int32. And in this case it has no problem doing so.
The result of the operation is 0.

It is only after this step has been completed that the assignment takes
place. That's an easy one: converting an int to a float. Mission
accomplished.

Martin

[Morten Wennevik]

As other people already mentioned, you need to tell the compiler to treat
the numbers as float.
Since they don't have a "." part they will be treated as integers.
A way to treat them as floats, without adding "." is:

float test = 55f/60f;

--
The hotmail account will most likely not be read, so please respond only
to the news group.

[Mohamoss]

Hi Dave
This happens because the clr assumes it is an integer operation, since
you are using two integer number (55 & 60).
All these will solve your problem
Double test = 55.0/60;
Double test = 55.0/60.0;
Or you explicitly tell the clr that you are not doing an int operation.
Double test = (double) 55/60;// explicit type conversion
Float test = (float) 55/60; // explicit type conversion
Hope this explains it
Regards,
Mohamed Mossad
ITworx on behalf of Microsoft GTSC Developer support for Middle East

[Lord Crc]

On Sun, 8 Feb 2004 01:26:26 +0100, "Martin Maat [EBL]"
<du***@somewhere.nl> wrote:

Not at all. Look at it this way: first the expression 55 / 60 is evaluated.
At that time the compiler doesn't know or care what you want to use the
result for.

However having separate integer and float division operators isn't
such a bad idea imho :) (ala basics / \ and pascals div /)

- Asbjørn