Develop Reference

Using 'continue' with modulo

I'm kinda new at c# programming so take it easy on me.
I couldn't find the answer to my (most likely) simple an stupid question (there's no stupid questions!!) so I post here.
I need to write a program which shows numbers from 1 to 10 that aren't divisble by 2, 3 and 8 using "continue" instruction.
My code:
static void Main(string[] args)
{
for (int i = 1; i <= 10; i++)
{
if (i % 2 == 0 && i % 3 == 0 && i % 8 == 0) continue;
Console.Write("{0} ", i);
}
Console.ReadKey();
}
It doesn't work, tho. The main issue is using &/&& operator. It should return both true and true. Help :(

I need to write a program which shows numbers from 1 to 10 that aren't divisble by 2, 3 and 8 using "continue" instruction.
The minimum number that can be divided by 8 with no remainder is 8. So the only numbers that could qualify are 8, 9 or 10.
if (8 % 2 == 0 && 8 % 3 == 0 && 8 % 8 == 0) // false
if (9 % 2 == 0 && 9 % 3 == 0 && 9 % 8 == 0) // false
if (10 % 2 == 0 && 10 % 3 == 0 && 10 % 8 == 0) // false
None of the numbers 8, 9, 10 can be divided by 2, 3, and 8 with a remainder of 0, so of course all numbers would be printed out, as continue would never be triggered. Are you sure it's "2, 3, AND 8" and not "2, 3, OR 8"?
if ((i % 2 == 0 && i % 3 == 0) || (i % 8 == 0)) continue;

IS there any way i can make this shorter

I am learning how to use untiy in my spare time by reading a beginners book and looking up stuff online in the book there is a exercise that asks me
to create a script that outputs the numbers from 1 to 10 in to the console but dose not output any multiple of 3 and 5 instead outputting the phrase "programming is awesome "
while i have achieved this task by using this code
using UnityEngine;
using System.Collections;
public class Loops : MonoBehaviour {
// Use this for initialization
void Start () {
for(int i = 1; i <= 10; i++)
{
if(i == 3 )
print ("Programming is Awesome!");
else if (i == 5)
print ("Programming is Awesome!");
else if (i == 6)
print ("Programming is Awesome!");
else if (i == 9)
print ("Programming is Awesome!");
else if (i == 10)
print ("Programming is Awesome!");
else
print (i);
}
}
}
i was wondering if there was any way to achieve the same result only by using less lines of code

You want to use the modulus (aka modulo) operator (%) for this task. It returns the remainder of a division, so when the result of a modulus operation is 0 you know you have a multiple of the divisor.
for (int i = 1; i <= 10; i++)
{
if(i % 3 == 0 || i % 5 == 0)
print("programming is awesome");
else
print(i);
}

for (int i = 1; i <= 10; i++)
{
print((i % 3 == 0 || i % 5 == 0)? "programming is awesome" : i));
}
check out also using ternary operator.

The point of this exercise is that you should calculate the multiples, not just make one condition for every value that you know is a multiple.
Use the modulo operator to check it a number is an even multiple of another. This shows what the modulo operator returns for some values:
i i % 3
------------
1 1
2 2
3 0
4 1
5 2
6 0
7 1
As you see, i % 3 evaluates to 0 when i is a multiple of three. You can use that to check if the number is a multiple of three:
if (i % 3 == 0) {
print ("Programming is Awesome!");
}
Now you should be able to do the same for five also, and incorporate it in your code.

Ways of doing this
if(i == 3 || i == 5 || i == 6 || i == 9 || i == 10){
print ("Programming is Awesome!");
}
else {
print (i);
}
Better way By Using the modulo operator
if( i % 3 == 0 || i % 5 == 0){
print ("Programming is Awesome!");
}
else {
print (i);
}
Use can also try
print((i % 3 == 0 || i % 5 == 0)? "Programming is Awesome!" : i));

Calculating a leap year without the leap year function

I need to calculate if the current year at the runtime of the program is a leap year (divisible by 4, not divisible by 100 but divisible by 400) but without using the DateTime.LeapYear property. Can anyone suggest anything?
//DateTimePicker code
private void dateTimePicker1_ValueChanged(object sender, EventArgs e)
{
DateTime now;
int[] months = {31,28,31,30,31,30,31,31,30,31,30,31};
now = DateTime.Now.Date;
if (now.Year / 4 == 0 && now.Year / 400 == 0)
{
months(1) = 29;
}
}

I think this covers the three criteria:
var year = now.Year;
if (year % 4 == 00 && !(year % 100 == 0 && year % 400 != 0))
{
....
}

Use the modulus operator % when checking divisibility. Also, when changing an array, use array indexers [], not parentheses:
if (now.Year % 4 == 0 && now.Year % 400 == 0)
{
months[1] = 29;
}

Shorten a DateTime condition check

if ((DateTime.Now.DayOfWeek != DayOfWeek.Friday && DateTime.Now.DayOfWeek != DayOfWeek.Saturday) &&
((DateTime.Now.Hour >= 10 && DateTime.Now.Hour < 13) || (DateTime.Now.Hour >= 20 && DateTime.Now.Hour < 23)))
I have to shorten this condition, any suggestions?

You could change the hours to use
(DateTime.Now.Hour % 12) +1 >= 10 && (DateTime.Now.Hour % 12) +1 < 13
Maybe even without the second check.
I don't think you can improve much more than that than looking for other methods like other answers
Update
I tested the above and its wrong, but this is more sadistic and works
var check = (DateTime.Now.Hours - 10 % 12) % 10;
var checkV = (DateTime.Now.Hours >= 10 && check < 3);
Test Code
for (int i = 0; i < 24; i++)
{
var check = (i - 10 % 12) % 10;
bool checkV = (i >= 10 && check < 3);
Console.WriteLine(i.ToString() + ": " + checkV.ToString());
}
Console.ReadKey();
Update 2
Complete shortened code
if( (int)DateTime.Now.DayOfWeek < 5 &&
DateTime.Now.Hours >= 10 &&
((DateTime.Now.Hours - 10 % 12) % 10) < 3)

Well, you could build an extension method:
public static bool BoundsCheck(this DateTime d, int min, int max, int min2, int max2)
{
return (d.DayOfWeek != DayOfWeek.Friday &&
d.DayOfWeek != DayOfWeek.Saturday &&
d.Hour >= min &&
d.Hour < max) ||
(d.Hour >= min2 && d.Hour < max2);
}
and then call it like this:
if (DateTime.Now.BoundsCheck(10, 13, 20, 23))...

Is this shorter? Maybe, but more important in my opinion it's more readable and maintainable:
var now = DateTime.Now;
var notAllowedDays = new[] { DayOfWeek.Friday, DayOfWeek.Saturday };
var allowedHours = Enumerable.Range(10, 3).Concat(Enumerable.Range(20, 3));
if(!notAllowedDays.Contains(now.DayOfWeek) && allowedHours.Contains(now.Hour))
{
}

if (!this.ItsPartyDay() && (this.ItsLunchTime() || this.ItsDinnerTime()))
{
...
}
private bool ItsPartyDay()
{
return (Int32)DateTime.Now.DayOfWeek >= 5;
}
private bool ItsLunchTime()
{
return (DateTime.Now.Hour >= 10 && DateTime.Now.Hour < 13);
}
private bool ItsDinnerTime()
{
return (DateTime.Now.Hour >= 20 && DateTime.Now.Hour < 23);
}

I don't think there is any reasonable solution but here a couple that come to mind. Use aliases for DateTime and DayOfWeek. One other option would be to assign all of those values to variables before the conditional.
So you could do things like;
string fri = DayOfWeek.Friday;
string sat = DayOfWeek.Saturday;
then use those in the conditional. Or;
using dt = DateTime;
Then you could do dt.Now.DayOfWeek
I personally would not recommend doing either of these things. You're not actually shortening the conditional, you're just refactoring. If you have a lot of these in one class it might be worth the trade off, otherwise it's probably not.
EDIT: The extension method suggestion by Michael Perrenoud is a reasonable solution that actually works really well.

Asp with C# Prime Number

I'm a 17 year old student currently in software engineering and web development and im having trouble right now with some of my coding. I need to make a project that will alow the user to input a number anywherefrom 0 to 999 and tell whether it is a prime number or not. The code i have so far is....
using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
public partial class _Default : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{
}
public void primeNumber()
{
int primeNumber1 = int.Parse(Request.Form["Text4"]);
if (primeNumber1 % 1 == 0 & ! (primeNumber1 % 2 == 0 &
primeNumber1 % 3 == 0 &
primeNumber1 % 4 == 0 &
primeNumber1 % 5 == 0 &
primeNumber1 % 6 == 0 &
primeNumber1 % 7 == 0 &
primeNumber1 % 8 == 0 &
primeNumber1 % 9 == 0))
{
Response.Write(" This is a prime number! ");
}
else
{
Response.Write(" This is not a prime Number! ");
}
}
}
... but i cannot get this program to display the correct answer. Any help would be greatly appreciated. Thanks!

You have got the concept of prime numbers wrong. Your code would for example report that 3 is not a prime number, because you check if the number divides evenly in three even if the number entered is three.
The simplest solution would be to loop from 2 and up to primeNumber1 - 1 and check if any of those divides evenly with the number. As you are using a loop, you also need a variable to hold what the result was, as you don't have a single expression that returns the result.
Something like:
bool prime = true;
for (int i = 2; i <= primeNumber1 - 1; i++) {
if (primeNumber1 % i == 0) {
prime = false;
}
}
This is of course the simplest possible solution that solves the problem, for reasonably small numbers. You can for example improve on the solution by exiting out of the loop as soon as you know that it's not a prime number.
You also don't need to loop all the way to primeNumber1 - 1, but only as high as the square root of the number, but you can find out about that if you read up on methods for checking prime numbers.
You need to handle the special cases of 1 and 2 also. By definition 1 is not a prime number, but 2 is.
http://en.wikipedia.org/wiki/Prime_number

bool IsPrime(int number) {
if (number == 1) return false;
if (number == 2) return true;
for (int i = 2; i < number; ++i) {
if (number % i == 0) return false;
}
return true;
}

A little google-fu or a little navel-gazing about prime numbers in general, will lead you to the naive algorithm:
For all n such that 0 < n:
There are two "special case" prime numbers, 1 and 2.
All even numbers > 2 are non-prime, by definition
If you think about the nature of factoring, the largest possible factor you have to consider is the square root of n, since above that point, the factors are reflexive (i.e., the possible factorizations of 100 are 1*100 , 2*50 , 4*25 , 5*20 , 10*10 , 20*5 , 25*4, 50*2 and 100*1 — and the square root of 100 is...10).
That should lead you to an implementation that looks something like this:
static bool IsPrime( int n )
{
if ( n < 1 ) throw new ArgumentOutOfRangeException("n") ;
bool isPrime = true ;
if ( n > 2 )
{
isPrime = ( 0 != n & 0x00000001 ) ; // eliminate all even numbers
if ( isPrime )
{
int limit = (int) Math.Sqrt(n) ;
for ( int i = 3 ; i <= limit && isPrime ; i += 2 )
{
isPrime = ( 0 != n % i ) ;
}
}
}
return isPrime ;
}

Anytime you find yourself in programming repeating a test on a sequential range of numbers you're doing the wrong thing. A better construct for this is a loop. This will give you the range of numbers in an identifier which can then be used to write the repetive code one time. For example I could rewrite this code
primeNumber1 % 2 == 0 &
primeNumber1 % 3 == 0 &
primeNumber1 % 4 == 0 &
primeNumber1 % 5 == 0 &
primeNumber1 % 6 == 0 &
primeNumber1 % 7 == 0 &
primeNumber1 % 8 == 0 &
primeNumber1 % 9 == 0))
As follows
bool anyFactors = false;
for (int i = 2; i <= 9; i++) {
if (primeNumber1 % i != 0) {
anyFactors = true;
break;
}
}
At this point I can now substitute the value allTrue for the original condition you wrote.
if (primeNumber1 % 1 == 0 && !anyFactors)
I can also expand the number of values tested here by substiting a different number for the conditional check of the loop. If I wanted to check 999 values I would instead write
for (int i = 2; i <= 999; i++) {
...
}
Additionally you don't want to use & in this scenario. That is for bit level and operations. You are looking for the logical and operator &&

Try the code below:
bool isPrimeNubmer(int n)
{
if (n >=0 && n < 4) //1, 2, 3 are prime numbers
return true;
else if (n % 2 == 0) //even numbers are not prime numbers
return false;
else
{
int j = 3;
int k = (n + 1) / 2 ;
while (j <= k)
{
if (n % j == 0)
return false;
j = j + 2;
}
return true;
}
}