How to display oddnumbers from a list of numbers(0-100)?

Question

How to display oddnumbers from a list of numbers(0-100) using two combo boxes and a button i.e( if we have two combo boxes placed on the form one containing numbers and then how to display even or odd numbers from a list of these numbers)

Thanks

Vulpes · Answer

You can also do this using LINQ: private void button1_Click(object sender, EventArgs e) { comboBox1.Items.AddRange(Enumerable.Range(0, 101).Where(i => i % 2 == 0).Cast ().ToArray() ); //even comboBox2.Items.AddRange(Enumerable.Range(0, 101).Where(i => i % 2 == 1).Cast ().ToArray()); // odd }

Miroslav Bucko · Answer

Here is an interesting solution //Because all even numbers are divisible by two, we can use the modulo division operator to see if there is any remainder when the //number is divided by two. If there is no remainder (it returns zero) then the number is definitely even. Zero is an even number; this //method is correct for zero. private void button1_Click(object sender, EventArgs e) { for (int i = 0; i { if (IsEven(i)) { combobox1.Items.Add(i); } if (IsOdd(i)) { combobox2.Items.Add(i); } } } private bool IsOdd(int value) { return value%2 == 1; } private bool IsEven(int value) { return value%2 == 0; }

Vilas Gite · Answer

Really Nice Explanation Sam & Vuples sir, I am trying to understand bitwise arithmetic "and" and "or".... Thanks for Great info.

Sam Hobbs · Answer

Yeah, I did not think about it enough. Doing an "and" with 1 is correct. For the benefit of beginners, it is useful to understand bitwise "and" and "or".

Vulpes · Answer

Yes, Sam's right that using bit arithmetic should be more efficient than the modulus operator when testing for oddness or evenness though, for non-negative numbers, you should be able to just AND with '1' to isolate the least significant bit: private void button1_Click(object sender, EventArgs e) { comboBox1.Items.AddRange(Enumerable.Range(0, 101).Where(i => (i & 1) == 0).Cast ().ToArray() ); //even comboBox2.Items.AddRange(Enumerable.Range(0, 101).Where(i => (i & 1) == 1).Cast ().ToArray()); // odd }

Sam Hobbs · Answer

Note that the expression "value % 2 != 0" uses division to determine the remainder. That is okay, but another way to do it is to use a bitwise "and". You just need to get an int value that has all bits on except the one we need; the low-order bit. So that value in hexadecimal would be all "F" except the low-order digit would be "E". An easy way to generate that value for an "int" is "Int32.MinValue + 1". So a variation of the IsOdd function from the article referenced above could be: public static bool IsOdd(int value) { //return value % 2 != 0; const int Flags = Int32.MinValue + 1; return (value & Flags) == 1; } Note that in that code, "Flags" is a const so it gets calculated once per execution (or once during compilation; Vulpes knows the truth). So Vulpes's LINQ code can be modified slightly to use the bitwise and operator instead of the modulus operator.

Vulpes · Answer

That's true, Vilas, but they are supported in .NET 3.5 (VS 2008) and the question has been asked in the .NET 3.0/3.5 category which is the only info we have.

Vilas Gite · Answer

@ Vulpes sir, LINQ features not support to VS2005. -------------------------------

Vilas Gite · Answer

@ Sapna.. Just open attachment and click on button... --------------------------------------------------------------------------------------------------- If this reply hepls your post…then " MARK AS ANSWER "!

Om prakash Singh · Answer

for odd numbers for (int i = 1; i { response.Write(i); } for Even numbers for (int i = 0; i { response.Write(i); }

Jiteendra Sampathirao · Answer

Hi. refer this: Link

How to display oddnumbers from a list of numbers(0-100)?

Answers (11)