Mail Archives: djgpp/2000/02/21/12:54:22
Thankyou so much!
The reason I posted it here is because you guys are all my DJGPP user
buddies! :P heh
--
-----------------------------------------------------
Go visit the United Fighters clan now!
http://unitedfighters.hypermart.net
David Cleaver <davidis AT ou DOT edu> wrote in message
news:38B09C3B DOT 58B9B2E7 AT ou DOT edu...
>
>
> Nite Hawk wrote:
> >
> > Ok so i'm a newbie but please help me out! Here is 2 problems I'm having
> > (god is this embarassing!)
> > 1. when I try to divide "taxPercent/100" it's value turns out to be "0"
(I'm
> > dividing a whole # such as 6 by 100 to get .06)
> > 2. How do I get a variable to contain a name? (I'm trying to get a name
as a
> > variable, is this possible?)
> >
> > I'll explain what I'm doing in #2 better:
> > I'm using...
> > int yourName;
> > but when the person enters his/her name it only shows the first letter
of
> > that persons name and also it will end the program without letting you
enter
> > the other variables in.
> >
> > Please help me!!!!!!
> > thankyou so very much and please do not laugh at my ignorence!
>
>
> Hey, we were all newbies once! I feel like a newbie myself, but I do
> know the answers to your questions so here we go:
>
> 1) When you write "taxPercent/100" you are doing an "integer-divide"
> operation. This means that only the integer value (0) is returned and
> the fractional part (.06) is discarded. So, you have a few options open
> to you for a solution:
> But first a question, What are you storing "taxPercent/100" into?
> The reason I ask is, if you are storing it in an "int" then you will
> never be able to get .06 stored there. You must use a float or a double
> variable to store "taxPercent/100". Now that you are storing into the
> correct variable, you have a couple of options open to you:
> for example, your program could look like this...
>
> int taxPercent = 6; file://this is your given value
> float myvariable = 0;
> myvariable = (float)taxPercent/(float)100;
>
> When you put "(float)" in front of a variable, that tells the program to
> try and treat the variable as a "float" type. This is called casting.
> In this example we are casting an int to a float. Float's are the
> variables that let you use decimal points, ie 0.06.
>
>
>
> 2) First off, as you know, int is not the correct type to hold a
> string. The varible that will hold strings is an array of characters.
> Strings and character arrays are basically the same thing. Here is an
> example of reading in a string from the command line (while the program
> is running):
>
> char name[40];
>
> scanf("%s", &name);
>
> The first line allocates memory for a variable that can hold 40
> characters, ie a string of 40 characters.
> The second line reads the your input and puts it into the variable
> "name".
>
> I hope this helps you in your endeavors. If you have any more questions
> please feel free to ask.
>
> Some people might tell you that this is the wrong group to post this
> kind of question in, and they'll probably be pretty blunt about it too.
> However, I'll tell you that usually questions relating to a specific
> language belong in comp.lang.c or whatever language it is you are asking
> about.
>
> Have fun programming!
>
> -David C.
- Raw text -