Message-ID:	<38B09C3B.58B9B2E7@ou.edu>
From:	David Cleaver <davidis AT ou DOT edu>
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Subject:	Re: Newbies need help too! <smile>
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Date:	Sun, 20 Feb 2000 20:00:27 -0600
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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;  //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.

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