Thread: Suggestions for how to improve upon this

You should state how does it not work.

That said, you are probably facing a problem due to expressions like this: ajay > (ram && shyam)
What you really mean to write is: ajay > ram && ajay > shyam.

The same goes for: ram == shyam == ajay.

Note that main returns int, always. Not nothing, as you have specified no return type.

You have to explicitly state your comparison on both sides of "&&"

You can't say:

Code:

value1 > (value2 && value3)

if what you really mean is

Code:

(value1 > value2) && (value1 > value3)

and the same for ==

Code:

value1 == value2 && value1 == value3

Originally Posted by newbcore

'm sorry, I don't understand what you mean here. Could you try to explain it differently?

Simply put the expression (ram && shyam) returns true (1) if both ram and shyam are non-zero, and false (0) if either or both of them are zero.

But with the expression ajay > (ram && shyam), what you really want to do is compare the value of ajay with the value of ram and the value of shyam. What you end up doing is comparing the value of ajay with 0 or 1, depending on the values of ram and shyam.

Originally Posted by newbcore

I also don't know what you mean as far as explaining how it works, because I thought you could look at the code and tell.

In this case the problem was obvious to me, but it usually helps if you state something like: "the code gives incorrect output. For example, I enter <such and such> and get <such and such> but I expected <such and such>".

First I would start with an optimal 3-input sorting network:

Code:

if (a < b) {
    if (b < c) {
        // a < b < c (case 1)
    } else if (a < c) {
        // a < c <= b (case 2)
    } else {
        // c <= a < b (case 3)
    }
} else {
    if (a < c) {
        // b <= a < c (case 4)
    } else if (b < c) {
        // b < c <= a (case 5)
    } else {
        // c <= b <= a (case 6)
    }
}

Then, inside cases 3, 4, and 6 you need to do some extra work to check if the first two are equal or not, and in the case of case 6 also checking if the last two are also equal, giving cases 3b, 4b, 6b, and 6c.