The Weird and The Wonderful

PIEBALDconsult wrote:

And how about removing the .ToString and .Equals ?

Chris Meech wrote:

The slower they work, the better.

if (node.Attributes["Publish"].InnerText.Trim().Contains(s))
                {
                    return;
                }

public static ArrayList NonTranslatable
        {
            get
            {
                nonTranslatable.Add("1st String");
                nonTranslatable.Add("2nd String");
                // ...
                nonTranslatable.Add("30th String");
                return nonTranslatable;
            }
        }

foreach (DictionaryEntry de in translationDictionary)
            {
                if (de.Key.ToString().Equals(node.Attributes["Publish"].InnerText.Trim()))
                    node.Attributes["Publish"].InnerText
                            = translationDictionary[node.Attributes["Publish"].InnerText.Trim()].ToString();
            }

Timothy Byrd wrote:

Simply putting an if (nonTranslatable.Count == 0) block around the code in the NonTranslatable property took the run time down from >75 minutes to ~30 seconds. (Fun fact: During the 75 minutes of processing, those repeated Add() calls caused the program to take up about 25MB of extra memory. That's a lot of strings to loop through and compare repeatedly. OMG )

Timothy Byrd wrote:

Not to mention all the calls to Trim() inside the loop. Or does the runtime optimize those out as invariant?

Megidolaon wrote:

What is the point of

if (node.Attributes["Publish"].InnerText.Trim().Contains(s))
{
    return;
}

A check for containting a single letter and then doing nothing?

Megidolaon wrote:

Maybe the XML contains trailing spaces?
Unless you trim them, the XML Nodes' InnerText wouldn't equal the DictionaryEntries with same

string xmlText = node.Attributes["Publish"].InnerText.Trim();

if (x == 0)
    {
        if (y != 0)
            {
                y = 0;
            }
    }
    else
    {
        if (y != x)
            {
                y = x;
            }
    }

{
  double x1,x2,z1,z2;

  x1= 1e-30/1e30;  /* x1 is positive zero */
  x2=-1e-30/1e30;  /* x2 is negative zero */
  z1=1/x1;         /* z1 is positive infinity */
  z2=1/x2;         /* z2 is negative infinity */
}

Rauhotz wrote:

I always do ...

switch (x)
{
case 0 : y = 0; break;
case 1 : y = 1; break;
case 2 : y = 2; break;
...
...
}

default: y = x;