Q. 1 Write algorithm for the following : a) to check whether an entered number is odd / even. b) to calculate sum of three numbers.
Code: int is_numeven(int number) { if(!(number % 2)) /* Mod return true; return false; } int SumThreeNumbers(int num1, int num2, int num3) { return num1+num2+num3; } The first example divides the given parameter by the smallest even number that can be divided with and then it checks the remainder. (correct me if I'm wrong) The second example should speak for itself.
I would suggest that producing someone's work for them is not in their best interests, when it comes to learning. It is, in fact, discouraged in the "Before you make a query" thread: It is better for the poster to post attempts and ask for help, or for the respondent to perhaps provide some pseudo-code or design guidelines. You may think otherwise, of course.
You're right, but the topic starter can ask questions after reading the small example. I mean, I just wrote a few lines of code for him and explained it a little bit. Not sure what I'm doing wrong, however I agree to your point that it might be better if I'd have just given a few guidelines or references to documents.
Also, please put code you write into code tags to preserve its indentation and formatting. I heartily recommend reading the "Before you make a query" thread, linked at the upper right of this page. There is nothing wrong with writing code or rewriting code -- provided that the learner is showing that he or she is putting forth effort to learn. Getting one's homework done entirely for free (or even for a price) is doing a disservice to the poster and to potential employers and coworkers who have to discover that the supposed skills were not actually learned.
I didn't look into it that way, I guess you're right. Thank you. (btw on the code tags, I saw I should have added them but couldn't edit my post - where is the edit button?)
I have done that for you and to edit your own posts you need to have minimum no of posts before you can do that.
// if even then return 1 else return 0 Code: int even_odd ( int num) { if (num & 1) return 0; return 1; }
just changing the way we acn also get the info from last bit Code: int is_numeven(int number) { if(!(number & 1)) return true; return false; }