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oops.. corrected!
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Ohh...look...I've really gone crazy with this now.
public static class Helper{
public static T Tee<T>(
this T @inVal,
Action<T> act){
act(@inVal);
return @inVal;
}
public static Byte[] GetBytes( this String @inVal){
Byte [] outBytes = new Byte[@inVal.Length];
int loopCount = 0;
foreach (Char c in @inVal){
outBytes[loopCount] = Convert.ToByte(c);
loopCount++;
}
return outBytes;
}
public static String DisplayBytes(this byte[] inBytes){
String outVal = String.Empty;
foreach (Byte b in inBytes){
outVal += Convert.ToString($"{b} ");
}
return outVal;
}
}
Try it like this and you get before and after again:
"What up!"
.Tee(Console.WriteLine)
.GetBytes()
.DisplayBytes()
.Tee(Console.WriteLine);
Output looks like:
What up!
87 104 97 116 32 117 112 33
Well, it's fun.
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It doesn't copy stdin to stdout. The only thing in common with "tee" is in the amount of parameters.
Bastard Programmer from Hell
If you can't read my code, try converting it here[^]
"If you just follow the bacon Eddy, wherever it leads you, then you won't have to think about politics." -- Some Bell.
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Yes, you are correct. I think the point that the author/presenter was attempting to make is that you can output the value and continue processing the value as input to yet another function.
That's why the author/presenter named the method Tee (which I hadn't seen before).
I looked it up and found the associated wiki article and just thought that was an interesting piece of history.
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raddevus wrote: I think the point that the author/presenter was attempting to make is that you can output the value and continue processing the value as input to yet another function. Like a decorator..
Bastard Programmer from Hell
If you can't read my code, try converting it here[^]
"If you just follow the bacon Eddy, wherever it leads you, then you won't have to think about politics." -- Some Bell.
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Eddy Vluggen wrote: Like a decorator..
cue Madonna ...touched for the very first time... I could not pass that up.
Ignoring the bad joke (if possible) I think your point is really interesting, because it is like a decorator.
Also, in an effort to completely beat this dead horse, how about the following addition?
If, nothing else, the added method has a great name : see SpaceOut.
public static class Helper{
public static T Tee<T>(
this T @inVal,
Action<T> act){
act(@inVal);
return @inVal;
}
public static Byte[] GetBytes( this String @inVal){
Byte [] outBytes = new Byte[@inVal.Length];
int loopCount = 0;
foreach (Char c in @inVal){
outBytes[loopCount] = Convert.ToByte(c);
loopCount++;
}
return outBytes;
}
public static String DisplayBytes(this byte[] inBytes){
String outVal = String.Empty;
foreach (Byte b in inBytes){
outVal += Convert.ToString($"{b:D3} ");
}
return outVal;
}
public static String SpaceOut(this string @inVal){
StringBuilder spacedItem = new StringBuilder();
foreach (Char c in @inVal){
spacedItem.Append($" {c} ");
}
Console.WriteLine(spacedItem.ToString());
return @inVal;
}
}
Now you can do this:
"What up!"
.Tee(Console.WriteLine)
.SpaceOut()
.GetBytes()
.DisplayBytes()
.Tee(Console.WriteLine);
And you will get the following:
What up!
W h a t u p !
087 104 097 116 032 117 112 033
Additionally interesting (or not) is that SpaceOut simply passes the input string along with no change since you only want the input to be printed with the extra spaces but don't want the output altered in this case.
I got a million of 'em!
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Nelek wrote: like a kid with brand shiny new shoes
No, unfortunately I have the same old shoes.
But, I did get a haircut recently, that we could talk about if you like.
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SpaceOut has a side-effect and is not functional by definition
It seems you're spacing out
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I thought a side effect would be to affect some property of the object??
I don't see where SpaceOut is be doing that.
Let me know what you're thinking so I can learn. thx
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In functional terms, writing to the console is a side-effect.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Ah, yes. That explains. thanks, very much.
However, I still kind of secretly like that SpaceOut method, but don't tell everyone because they'll think I'm not all functional-minded.
Seriously though, that was good to learn because it makes me think the right way about functional programming that it should be just like a mathematical expression. Do the thing, return the value.
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I can recommend learning Haskell, a pure functional language.
In Haskell all I/O is considered a side-effect (database actions, drawing, printing to console, writing a file, etc.).
If you're doing any I/O in Haskell your function must return the special System.IO object.
Here's a nice example of Hello World in Haskell[^].
I don't know Haskell very well myself, just followed a course at University, but it certainly changed how I think about my code and it taught me some nice functional concepts
My favorite example of how nice, readable and succinct a functional programming and Haskell can be is an implementation of the quick sort algorithm:
qsort [] = []
qsort (x: xs) = qsort smaller ++ [x] ++ qsort bigger
where
smaller = [a | a <- xs, a <= x]
bigger = [a | a <- xs, a > x]
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Quote:
public static Byte[] GetBytes( this String @inVal){
Byte [] outBytes = new Byte[@inVal.Length];
int loopCount = 0;
foreach (Char c in @inVal){
outBytes[loopCount] = Convert.ToByte(c);
loopCount++;
}
return outBytes;
}
That should probably be:
public static Byte[] GetBytes( this String @inVal ) => System.Text.Encoding.UTF8.GetBytes(@inVal);
Remember, some characters aren't single bytes.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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Richard Deeming wrote: Remember, some characters aren't single bytes
Very good.
A very amateur mistake by me.
I was excited about the idea and didn't think about the code that is already there for me.
Lame!
Thanks again!
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Why is an @ sign being used for @inVal? Aren't those only used for naming a variable after a reserved keyword?
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icemanind wrote: Why is an @ sign being used for @inVal?
I wondered the same thing.
This was the choice of the author/presenter and I hadn't seen that before either.
He actually had his param named @this so it was even worse to me in that way.
I didn't know C# allowed a var to be named with a beginning @ sign either, but apparently you can.
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I knew you could have an @ sign in it. And it makes sense for @this because "this" is a reserved keyword in C#. Without the @ sign, naming a variable "this" would cause an error. But there is no keyword called "inVar", which is why I was confused about that one.
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Honestly, it was me being ignorant.
However, you've made me think about this and I get the point now.
It sounds like I'm being sarcastic but I'm really not.
I know this thread has been odd but it has really made me rethink and firstThink a number of things.
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Hi, Raddevus,
Really enjoying this discussion !
It would be interesting if you could, somehow:
public static T TeeHee<T>(this T tee, params Func<T, T>[] funcs)
{
foreach (var func in funcs)
{
}
return tee;
} But, the obvious problem is that all the funcs have to have the same return type.
cheers, Bill
«... thank the gods that they have made you superior to those events which they have not placed within your own control, rendered you accountable for that only which is within you own control For what, then, have they made you responsible? For that which is alone in your own power—a right use of things as they appear.» Discourses of Epictetus Book I:12
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Func<T, TResult> is covariant[^] on the return type, so you could pass in functions which returned a more derived type than the declared return type.
public static T TeeHee<T>(this T tee, params Func<T, object>[] funcs)
{
foreach (var func in funcs)
{
object x = func(tee);
}
return tee;
}
42.TeeHee(
i => i,
i => $"The answer is {i}",
i => new Answer(i)
);
The more important question would be, what are you intending to do with the returned values?
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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