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?
9  1101 
"Christopher Weaver" <we*****@verizon.net> wrote in
news:uT*************@TK2MSFTNGP15.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 de***@sio.NOSPAMmidco.net
Starting Points: http://sio.midco.net/denro/www
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" <de***@NOSPsio.midco.net> wrote in message
news:Xn***************************@216.196.97.142. .. "Christopher Weaver" <we*****@verizon.net> wrote in
news:uT*************@TK2MSFTNGP15.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
de***@sio.NOSPAMmidco.net
Starting Points:
http://sio.midco.net/denro/www
Christopher Weaver wrote: 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?
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-
Personally I prefer to use:
o = c * 0.01;
Is it faster?
Perharps negligible.
"Christopher Weaver" <we*****@verizon.net> wrote in message
news:ej**************@tk2msftngp13.phx.gbl... 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" <de***@NOSPsio.midco.net> wrote in message
news:Xn***************************@216.196.97.142. .. "Christopher Weaver" <we*****@verizon.net> wrote in
news:uT*************@TK2MSFTNGP15.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
de***@sio.NOSPAMmidco.net
Starting Points:
http://sio.midco.net/denro/www
Good point. It's so nice to have a group that pays attention and cares
enough to write.
"Rick Lones" <Wr******@YcharterZ.net> wrote in message
news:TO*****************@fe05.lga... Christopher Weaver wrote: 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?
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-
I think I like it better too.
"Altramagnus" <al*********@hotmail.com> wrote in message
news:42********@news.starhub.net.sg... Personally I prefer to use:
o = c * 0.01;
Is it faster?
Perharps negligible.
"Christopher Weaver" <we*****@verizon.net> wrote in message
news:ej**************@tk2msftngp13.phx.gbl... 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" <de***@NOSPsio.midco.net> wrote in message
news:Xn***************************@216.196.97.142. .. "Christopher Weaver" <we*****@verizon.net> wrote in
news:uT*************@TK2MSFTNGP15.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
de***@sio.NOSPAMmidco.net
Starting Points:
http://sio.midco.net/denro/www
Altramagnus <al*********@hotmail.com> wrote: Personally I prefer to use:
o = c * 0.01;
Is it faster?
Perharps negligible.
I don't think it's nearly as clear at a glance that that's dividing by
a hundred though. Just MHO.
--
Jon Skeet - <sk***@pobox.com> http://www.pobox.com/~skeet
If replying to the group, please do not mail me too
i agree, but you can put a comment or something.
"Jon Skeet [C# MVP]" <sk***@pobox.com> wrote in message
news:MP************************@msnews.microsoft.c om... Altramagnus <al*********@hotmail.com> wrote: Personally I prefer to use:
o = c * 0.01;
Is it faster?
Perharps negligible.
I don't think it's nearly as clear at a glance that that's dividing by
a hundred though. Just MHO.
--
Jon Skeet - <sk***@pobox.com>
http://www.pobox.com/~skeet
If replying to the group, please do not mail me too
Altramagnus <al*********@hotmail.com> wrote: i agree, but you can put a comment or something.
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 - <sk***@pobox.com> http://www.pobox.com/~skeet
If replying to the group, please do not mail me too This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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