Answer wont display right

JR · **Posted:** Tue Oct 19, 2004 10:27 pm

this program calculates the points the user got from the questions, well a part of it. now when i putput the points i get some strange number which is wrong, how do i make it output the right nunbmer at the end of the 5 questions when it calculated all the points?

code:

// Alex R
// A5
// October 10th

#include <iostream.h>
#include <stdlib.h>
#include <time.h>

/*
Name: Ran
Purpose: Generate a random number
Gets: A int variable
Retrun: num which is the random number

*/
int ran(int &num)
{

num= (rand() % 100)+1; //Gets a random number from 1-100
return (num);
}

/*
Name: add
Purpose: adds 2 generated random numbers
Gets: 2 numbers(variables)
Retrun: the addition between the numbers

*/
int add (int &a, int &b)
{
int total=0;
ran(a);
ran(b);
total = a + b;
return (a+b);
}

/*
Name: subtract
Purpose: subtracts 2 generated random numbers
Gets: 2 variables
Retrun: subtraction of 2 numbers

*/
int subtract (int &a, int &b)
{
int total=0;
ran(a);
ran(b);
total = a - b;
return (a-b);
}

/*
Name: multiply
Purpose: multiplies 2 generated random numbers
Gets: 2 variables
Retrun: multiplication of 2 numbers

*/
int multiply (int &a, int &b)
{
int total=0;
ran(a);
ran(b);
total = a * b;
return (a*b);
}

/*
Name: answers
Purpose: checks the answers
Gets: one variable
Retrun: N/A

*/
void answers( int &total){

// 10 points if answered correctly the first time.
// 5 points if answered correctly the second time.
// 2 points three times a charm.
// 0 points if you struck out!

int answer;
int wrong=0;
int points;

for(int i=0; i < 3; i++){

cin>>answer;

if( answer != total ){
if (i < 2){
cout<<"WRONG, Try again."<<endl;
wrong++;
}
if( i == 2){
cout<<"You have missed 3 times. The answer is "<<total<<endl;
points+=0;
}
}else if ( answer == total ){

cout<<"CORRECT"<<endl;

if (wrong == 0 ){
points+=10;
}else if ( wrong == 1){
points+=5;
}else if (wrong == 2){
points+=2;

}
break;

}

}

cout<<"You have "<<points<<" points \n";

}

/*
Name: menu
Purpose: Outputs the menu
Gets: 2 variables
Retrun: N/A

*/
void Menu(int &a, int &b )
{
//int x;
int total, choice;

cout<<endl;
cout<<"Chose one of the following.\n";
cout<<"1. Addition\n2. Subtraction\n3. Multiplication\n4. Exit\n";
cin>>choice;
switch (choice){

case 1: add(a, b);
cout<<a<<" + "<<b<<" = "<<endl;
total=a+b;
answers(total);
break;

case 2: subtract(a, b);
cout<<a<<" - "<<b<<" = "<<endl;
total=a-b;
answers(total);
break;

case 3: multiply(a, b);
cout<<a<<" * "<<b<<" = "<<endl;
total=a*b;
answers(total);
break;

case 4: cout<<"Bye!!!" ;
break;

default: cout<<"The options are only 1-4.";
break;
}
}

int main ()
{
srand(time(0));

int a, b;

// We can only have 5 problems.
for ( int i=0; i < 5; i++){
Menu( a, b );
}

cin.get();
cin.get();

return 0;
}

Andy · **Posted:** Wed Oct 20, 2004 3:02 pm

u have to define points to 0... if u dont, the program will just give you what ever is at that specific memory allocation at that time

JR · **Posted:** Thu Oct 21, 2004 6:14 am

lol i misseds that one thx

JHanson90 · **Posted:** Fri Oct 22, 2004 11:20 pm

Since the title of this topic applies to the problem that I'm having, I might as well post this in here.

I'm learning C (note for the careless reader: not C++), and I'm in the second chapter of a book about it. My exercise is to write a program that will ask the user for a number, and the program returns all of that number's factors. It compiles fine. But at run-time, I encounter an error that reads "Floating point exception" when it gets to the 'for' loop part. I tried to Google it, but I couldn't understand much of what was said being a noob to C. Here's the program:

code:

/* Chapter 2: Section 4, Exercise 3 */
#include <stdio.h>

int main (void)
{
/* Tell the user what this program does. */
printf ("This program will first ask you for an integer. It will then give you all of the factors of that integer.\n");

/*
   Ask for the integer, and put it into the input variable.
*/
int input;
printf ("Please input any integer: ");
scanf ("%d", &input);

int i;
printf ("Here are the factors of your input (%d).\n", input);
for (i = 0; i < input; i++)
{
if (input / i) /* If variable input divided by variable i is any other number than 0, execute the following. */
{
/*
   If there are no more factors left to bother with , use a period instead of another comma
   (e.g. it's pointless to try to see if 20 / 11 will work, because I know there won't be any
   factors larger than half of the input).
*/
if ((i * 2) > input)
printf ("%d.\n", i);
else {
printf ("%d, ", i);
}
}
}

return 0;
}

I tried replacing the '(input / i)' with '(input % i)', and a few other things, just to see if it would at least fix the error regardless of the task at hand, but it always gave me the runtime error unless I commented out that if statement.

wtd · **Posted:** Sat Oct 23, 2004 3:35 am

Which compiler are you using?

code:

for (i = 0; i < input; i++)

When you divide by i, the first time through the loop, you're dividing by zero. Division by zero does bad bad things.

Since 0 isn't a factor of anything, I suggest starting with:

code:

for (i = 1; i < input; i++)

And a few other modifications:

code:

#include <stdio.h>

int main (void)
{
int input = 0, i;

fprintf(stdout, "This program will first ask you for an integer. It will then give you all of the factors of that integer.\n");

fprintf(stdout, "Please input any integer: ");
fscanf(stdin, "%d", &input);

fprintf(stdout, "Here are the factors of your input (%d).\n", input);
for (i = 1; i <= input; i++)
{
if (input % i == 0)
{
if (i == input)
printf("%d.\n", i);
else
printf("%d, ", i);
}
}

return 0;
}

Andy · **Posted:** Sat Oct 23, 2004 9:55 am

yea thats the problem.. btw, wtd what does %d do? im not too familiar wit C.. only c++

wtd · **Posted:** Sat Oct 23, 2004 2:14 pm

%d is a format specifier.

http://www.dgp.utoronto.ca/~ajr/209/notes/printf.html

Andy · **Posted:** Sat Oct 23, 2004 5:11 pm

OOooo... thats nice.. thx man

wtd · **Posted:** Sat Oct 23, 2004 5:50 pm

Conveniently, those standards are implemented by many languages.

O'Caml:

code:

open Printf
include Printf

let a = 9 in
printf "%d\n" a

Java 1.5/5.0:

code:

import java.lang.*;
import java.io.*;

public class PrintfTest {
public static void main(String[] args) {
final int A = 9;
System.out.printf("%d\n", A);
}
}

Perl:

code:

my $a = 9;
printf "%d\n", $a;

Ruby:

code:

a = 9
printf "%d\n", a

JHanson90 · **Posted:** Sat Oct 23, 2004 11:00 pm

wtd wrote:

What compiler are you using?

The GNU Compiler Collection (commonly abbreviated as "GCC").

wtd wrote:

When you divide by i, the first time through the loop, you're dividing by zero. Division by zero does bad bad things.

Ah, so really just a math error (dumb of me not to remember that you can't divide by 0). Thanks.

wtd wrote:

And a few other modifications:

code:

#include <stdio.h>

int main (void)
{
int input = 0, i;

fprintf(stdout, "This program will first ask you for an integer. It will then give you all of the factors of that integer.\n");

fprintf(stdout, "Please input any integer: ");
fscanf(stdin, "%d", &input);

fprintf(stdout, "Here are the factors of your input (%d).\n", input);
for (i = 1; i <= input; i++)
{
if (input % i == 0)
{
if (i == input)
printf("%d.\n", i);
else
printf("%d, ", i);
}
}

return 0;
}

Hmm, are fprintf() and fscanf() more efficient or something?