Question

Hey everyone, I'm looking for some help with a couple of problems in a code I've been working on! Any help would be greatly appreciated! The first thing is that when it goes to report the min/max/etc. stuff (this will make sense in the picture), in a situation where I don't enter any odd numbers (or even), nothing should be reported, but now all I'm getting is an entry of 0 in that situation, and I've really no clue how to fix that. The second issue is that whenever I go to actually enter the integers, regardless of how many I tell it to use, it always stops when I get to 1 before that. Cont..

Answer 1

So for example when I say I want to enter 4 integers, it will stop me and move on after only 3, and it's always like that. Attached is an image of what it should look like (though I mentioned what I need to do about there being no odd/even numbers). And here is the code itself:

// Assignment 2.cpp : main project file.

#include <StdAfx.h>
#include <iostream>
#include <string>
using namespace std;
int main(void)
{

//Declaring
int min=0,max=0,sumo=0,sume=0,n,j=0,totalnum=0,ne=0,no=0;


//Creating a cout statement so the user can input the number of integers the program will compile
cout << "How many integers do you want to enter > ";
cin >> totalnum;

if(totalnum<=0)//This is to verify that a positive integer was entered
{
cout <<"Please enter a number at least greater than 1 ";
cin>> totalnum;
}
else
{
do
{
cout << "Enter integer #"<< j++ << " ";
cin >> n;
if(j==0)
{
min=max=n;
j+1;
}
else
{
if(n>max)
{
max=n;
}
if(n<min)
{
min=n;
}
}

if(n%2==0)
{
sume+=n;
ne++;
}
else
{
sumo+=n;
no++;
}
}while(j+1<totalnum);
}
cout << "Max is " << max << "\n";
cout << "Min is " << min << "\n";
cout << "Number of odds is " << no << "\n";
cout << "Number of evens is " << ne << "\n";
cout << "Sum of odds is " << sumo << "\n";
cout << "Sum of evens is " << sume << "\n";
cout << "Integer average is " << (sume +sumo)/totalnum << "\n";
return system("pause");
}

Answer 2

1. The way you've stuctured your code, you will always have some kind of output at the end (printint out the max, min, sum, etc.). Also, if you enter 0 for totalnum, you get a division by zero error (since the bottom code always executes and you are comptuing the mean using totalnum).

2. At the start of the program code, you initialize j to 0. At the beginning of the do-while loop, you increment j (it goes from [1,2,3] if totalnum = 4), so if totalnum = 4, j+1 will go from [2,3,4] before breaking out of the loop; it will only execute 3 times instead of the expected 4 :(

Number 2 is fixed if you post-increment j within the while() condition at the end: http://ideone.com/Oud8w

To solve number 1, use logical strucutres in your code (if statements, loops, etc) to check the input; if you don't want the final part of your code to execute if no (# of odd numbers is 0) or if a bad totalnum was input, then you can use something like

if (no != 0 && ne != 0 && totalnum > 0) // if there is no bad input, output the final result
{
// output stuff
}

http://ideone.com/20Ixg

Answer 3

First, wouldn't it be easier to use a for loop to enter your integers rather than incrementing a counter and performing a check? Example (in pseudo code).

for c = 1 to totalnum;
print "Enter integer # " c;
input n

Answer 4

@jagatuba

I thought the do {} while () part was already doing that :-P

Answer 5

ahh.... right.

Answer 6

the for loop is indeed better to use:

http://ideone.com/gb13h

Answer 7

Yes, but it doesn't seem that efficient to me, but I'm not a C expert like yourself. Also why not use an int array to store the entries. That way all entries are entered and then calculations can occur afer the entry loop.

Answer 8

Yes I would prefer the for loop in this instance.

Answer 9

Also, why only positive integers? the screenshot show both positive and negative integers.

Answer 10

Slight alteration to some of your code ag. http://ideone.com/vpWzt

Answer 11

right; only totalnum needs to be positive

Answer 12

Just formatting. Change it so it asks for #1 #2 etc instead of #0 #1 etc
and made Max prin to next line.

Answer 13

Oh right DUH.

Answer 14

Also where is min and max of odd and even; and average of od and even?

Answer 15

This is closer to the output of the screen shot: http://ideone.com/aaXrk

Answer 16

Remove min, max, and integer average display and it will be identical to the screen shot.