From: "Perhe Virolainen" <matttt AT nic DOT fi>
Newsgroups: comp.os.msdos.djgpp
References: <88q38n$6cf$1 AT news3 DOT infoave DOT net>
Subject: Re: Newbies need help too! <smile>
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Date: Sat, 26 Feb 2000 21:56:50 +0200
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Organization: NIC Tietoverkot Oy - NIC Data Networks Ltd.
To: djgpp AT delorie DOT com
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Reply-To: djgpp AT delorie DOT com


Nite Hawk <malcolm3 AT bulloch DOT net> wrote in message
:88q38n$6cf$1 AT news3 DOT infoave DOT net...
> 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!

I will not laugh.
Answer for problem 1:
taxPercent/100
Are you sure that tax percent is not an integrer (I'm quite sure that
taxPercent is not over 100). Integers CAN NOT
hold any decimals. Use double or float.
Answer for problem 2:
If you want to store strings use char[];
like this for C++:
#include<iostream.h>

int main()
{
char name[50]; // Remeber to hold enough room to '\0'
cout<<" Please enter your name\n";
cin >> name;
cout << "\nYour name is: "<<name;
return 0;
}

That \n causes next line to be used
Or like this for C:
#include<stdio.h>

void main(void)
{
char name[50]; /* Remeber to hold enough room to '\0' */
printf("Please enter your name\n");
gets(name);
printf("\nYour name is: %s",name);
}