Really Simple Math

[Christopher Weaver]

I've declared o to be a double and c an int. No matter what the value of c,
o is never greater than 0 after the following line of code:

o = (c / 100);


Any ideas why?

[Dennis Roark]

Re: Really Simple Math

"Christophe r Weaver" <we.aver@verizo n.net> wrote in
news:uTS#76LYFH A.584@TK2MSFTNG P15.phx.gbl:
[color=blue]
> I've declared o to be a double and c an int. No matter what the value
> of c, o is never greater than 0 after the following line of code:
>
> o = (c / 100);
>
>
> Any ideas why?
>
>
>[/color]
c/100 performs integer division. If c is less than 100 then c/100 will
be a 0. If you want the value you expect, use

o = c / 100.0 now you do floating point division.
Any introductory programming book will go over this.

(BTW, o is a poor name of a variable of 1 character length.)


--
Dennis Roark

denro@sio.NOSPA Mmidco.net
Starting Points:

http://sio.midco.net/denro/www

[Christopher Weaver]

Re: Really Simple Math

Thanks. I knew it was basic. In fact, as I read your reply I recalled the
point being made in my first semester c++ class. Years of Pascal has erased
much of that.

BTW, the 'o' was included for debugging purposes only.


"Dennis Roark" <denro@NOSPsio. midco.net> wrote in message
news:Xns9660C60 5932A7denromidc onet@216.196.97 .142...[color=blue]
> "Christophe r Weaver" <we.aver@verizo n.net> wrote in
> news:uTS#76LYFH A.584@TK2MSFTNG P15.phx.gbl:
>[color=green]
>> I've declared o to be a double and c an int. No matter what the value
>> of c, o is never greater than 0 after the following line of code:
>>
>> o = (c / 100);
>>
>>
>> Any ideas why?
>>
>>
>>[/color]
> c/100 performs integer division. If c is less than 100 then c/100 will
> be a 0. If you want the value you expect, use
>
> o = c / 100.0 now you do floating point division.
> Any introductory programming book will go over this.
>
> (BTW, o is a poor name of a variable of 1 character length.)
>
>
> --
> Dennis Roark
>
> denro@sio.NOSPA Mmidco.net
> Starting Points:
> http://sio.midco.net/denro/www[/color]

[Rick Lones]

Re: Really Simple Math

Christopher Weaver wrote:[color=blue]
> I've declared o to be a double and c an int. No matter what the value of c,
> o is never greater than 0 after the following line of code:
>
> o = (c / 100);
>
>
> Any ideas why?
>
>[/color]
No idea IF your claim is correct. However, I think what you really mean is "No
matter what the value of c (as long as it's less than 100), o is never greater
than 0."

The expression (c / 100) is apparently calculated using integer arithmetic,
which is perfectly reasonable since c and 100 are both integer expressions. The
integer result of this division will be zero when c is less than 100. I suggest
you test your "no matter what" claim with some larger values of c (e.g., 100,
101, 199, 200, 201, 299, etc.) and watch what happens.

-rick-

[Altramagnus]

Re: Really Simple Math

Personally I prefer to use:

o = c * 0.01;

Is it faster?
Perharps negligible.


"Christophe r Weaver" <we.aver@verizo n.net> wrote in message
news:ej27AHMYFH A.1092@tk2msftn gp13.phx.gbl...[color=blue]
> Thanks. I knew it was basic. In fact, as I read your reply I recalled
> the point being made in my first semester c++ class. Years of Pascal has
> erased much of that.
>
> BTW, the 'o' was included for debugging purposes only.
>
>
> "Dennis Roark" <denro@NOSPsio. midco.net> wrote in message
> news:Xns9660C60 5932A7denromidc onet@216.196.97 .142...[color=green]
>> "Christophe r Weaver" <we.aver@verizo n.net> wrote in
>> news:uTS#76LYFH A.584@TK2MSFTNG P15.phx.gbl:
>>[color=darkred]
>>> I've declared o to be a double and c an int. No matter what the value
>>> of c, o is never greater than 0 after the following line of code:
>>>
>>> o = (c / 100);
>>>
>>>
>>> Any ideas why?
>>>
>>>
>>>[/color]
>> c/100 performs integer division. If c is less than 100 then c/100 will
>> be a 0. If you want the value you expect, use
>>
>> o = c / 100.0 now you do floating point division.
>> Any introductory programming book will go over this.
>>
>> (BTW, o is a poor name of a variable of 1 character length.)
>>
>>
>> --
>> Dennis Roark
>>
>> denro@sio.NOSPA Mmidco.net
>> Starting Points:
>> http://sio.midco.net/denro/www[/color]
>
>[/color]

[Christopher Weaver]

Re: Really Simple Math

Good point. It's so nice to have a group that pays attention and cares
enough to write.


"Rick Lones" <WrlonesX@Ychar terZ.net> wrote in message
news:TOPke.1038 2$eR.8486@fe05. lga...[color=blue]
> Christopher Weaver wrote:[color=green]
>> I've declared o to be a double and c an int. No matter what the value of
>> c, o is never greater than 0 after the following line of code:
>>
>> o = (c / 100);
>>
>>
>> Any ideas why?[/color]
> No idea IF your claim is correct. However, I think what you really mean
> is "No matter what the value of c (as long as it's less than 100), o is
> never greater than 0."
>
> The expression (c / 100) is apparently calculated using integer
> arithmetic, which is perfectly reasonable since c and 100 are both integer
> expressions. The integer result of this division will be zero when c is
> less than 100. I suggest you test your "no matter what" claim with some
> larger values of c (e.g., 100, 101, 199, 200, 201, 299, etc.) and watch
> what happens.
>
> -rick-[/color]

[Christopher Weaver]

Re: Really Simple Math

I think I like it better too.


"Altramagnu s" <altramagnus@ho tmail.com> wrote in message
news:4293d9bf$1 @news.starhub.n et.sg...[color=blue]
> Personally I prefer to use:
>
> o = c * 0.01;
>
> Is it faster?
> Perharps negligible.
>
>
> "Christophe r Weaver" <we.aver@verizo n.net> wrote in message
> news:ej27AHMYFH A.1092@tk2msftn gp13.phx.gbl...[color=green]
>> Thanks. I knew it was basic. In fact, as I read your reply I recalled
>> the point being made in my first semester c++ class. Years of Pascal has
>> erased much of that.
>>
>> BTW, the 'o' was included for debugging purposes only.
>>
>>
>> "Dennis Roark" <denro@NOSPsio. midco.net> wrote in message
>> news:Xns9660C60 5932A7denromidc onet@216.196.97 .142...[color=darkred]
>>> "Christophe r Weaver" <we.aver@verizo n.net> wrote in
>>> news:uTS#76LYFH A.584@TK2MSFTNG P15.phx.gbl:
>>>
>>>> I've declared o to be a double and c an int. No matter what the value
>>>> of c, o is never greater than 0 after the following line of code:
>>>>
>>>> o = (c / 100);
>>>>
>>>>
>>>> Any ideas why?
>>>>
>>>>
>>>>
>>> c/100 performs integer division. If c is less than 100 then c/100 will
>>> be a 0. If you want the value you expect, use
>>>
>>> o = c / 100.0 now you do floating point division.
>>> Any introductory programming book will go over this.
>>>
>>> (BTW, o is a poor name of a variable of 1 character length.)
>>>
>>>
>>> --
>>> Dennis Roark
>>>
>>> denro@sio.NOSPA Mmidco.net
>>> Starting Points:
>>> http://sio.midco.net/denro/www[/color]
>>
>>[/color]
>
>[/color]

[Jon Skeet [C# MVP]]

Re: Really Simple Math

Altramagnus <altramagnus@ho tmail.com> wrote:[color=blue]
> Personally I prefer to use:
>
> o = c * 0.01;
>
> Is it faster?
> Perharps negligible.[/color]

I don't think it's nearly as clear at a glance that that's dividing by
a hundred though. Just MHO.

--
Jon Skeet - <skeet@pobox.co m>

http://www.pobox.com/~skeet

If replying to the group, please do not mail me too

[Altramagnus]

Re: Really Simple Math

i agree, but you can put a comment or something.


"Jon Skeet [C# MVP]" <skeet@pobox.co m> wrote in message
news:MPG.1cfe29 b83816ee8098c1b e@msnews.micros oft.com...[color=blue]
> Altramagnus <altramagnus@ho tmail.com> wrote:[color=green]
>> Personally I prefer to use:
>>
>> o = c * 0.01;
>>
>> Is it faster?
>> Perharps negligible.[/color]
>
> I don't think it's nearly as clear at a glance that that's dividing by
> a hundred though. Just MHO.
>
> --
> Jon Skeet - <skeet@pobox.co m>
> http://www.pobox.com/~skeet
> If replying to the group, please do not mail me too[/color]

[Jon Skeet [C# MVP]]

Re: Really Simple Math

Altramagnus <altramagnus@ho tmail.com> wrote:[color=blue]
> i agree, but you can put a comment or something.[/color]

If you have to put a comment on one version to make it clearer, and you
haven't proved it to be faster, why do you prefer it? I'd take
readability over unproven speed any day.

--
Jon Skeet - <skeet@pobox.co m>

http://www.pobox.com/~skeet

If replying to the group, please do not mail me too