for i := X to Y inclusive... how?

[Willicious 8]

If you want to specify "i" as being any value between 0 - 7, you can use

for i := 0 to 7 do

And, as I understand it, this will run code as long as "i" matches any value between 0 and 7.

 

But, what if you only want to run the code when "i" matches every value between 0 and 7 at the same time?

 

So, for example, rather than typing:

if (X, Y - 0) and (X, Y - 1) and (X, Y - 2) and (X, Y - 3) etc...

 

Is there a way to express it as (X, Y - i) where i can be any value between 0 and 7, but every value must be accounted for simultaneously?

[Willicious 8]

Replying so I can follow the topic

[dummzeuch 1505]

So, what is (x, y-0) etc. meant to evaluate to? That's not a Boolean expression nor an integer that could be processed with a bit manipulation.

 

Ok, assuming this somehow is a Boolean anyway, that's the code I'd use:

 

Res := True;

For i := 0 to 7 do begin

  If not (x, y-i) then begin

    Res := False;

    Break;

  End;

End;

if Res then

  ....

[zed 14]

It's hard to understand what exactly you want. How about this:

 

if (X in [0..7]) and (Y in [0..7]) then begin

...

end

 

Edited May 7, 2023 by zed

[Attila Kovacs 629]

if (i >=0) and (i <=7) then do_something;

[PeterBelow 238]

5 hours ago, Willicious said:

But, what if you only want to run the code when "i" matches every value between 0 and 7 at the same time?

That makes no sense, a variable can only have one value at a time. Of course it may be a complex "value", like an array or set, but that does not seem to be what you want.

[Willicious 8]

6 hours ago, zed said:

 

if (X in [0..7]) and (Y in [0..7]) then

 

 

6 hours ago, Attila Kovacs said:

if (i >=0) and (i <=7) then do_something;

 

The above suggestions will run the code if i is equal to any number between 1 and 7. So, it's more like this:

if (i = 1) or (i = 2) or (i = 3) or (i = 4) etc...

 

What I need is for the code to run only when all numbers are satisfied, like this (note "and" instead of "or"😞

if (i - 1) and (i = 2) and (i = 3) and (i = 4) etc...

 

I realise that, of course, it's not possible for i to equal more than one value at once, that's why I need help with this. Essentially, it's a collision detection check in which the character has to turn around if they reach a vertical wall of at least 7 pixels in height, but  - all 7 pixels have to be filled. The following code works, but I want to try and simplify it if possible:

if HasPixelAt(X, Y) and HasPixelAt(X, Y -1)
  and HasPixelAt(X, Y -2) and HasPixelAt(X, Y -3)
  and HasPixelAt(X, Y -4) and HasPixelAt(X, Y -5)
  and HasPixelAt(X, Y -6) and HasPixelAt(X, Y -7) then
begin
...
end;

 

Thanks in advance, I hope this explains the question a little better.

[zed 14]

Here is the answer:

 

[PeterBelow 238]

4 minutes ago, Willicious said:

 

 

The above suggestions will run the code if i is equal to any number between 1 and 7. So, it's more like this:

if (i = 1) or (i = 2) or (i = 3) or (i = 4) etc...

 

What I need is for the code to run only when all numbers are satisfied, like this (note "and" instead of "or"😞

if (i - 1) and (i = 2) and (i = 3) and (i = 4) etc...

 

I realise that, of course, it's not possible for i to equal more than one value at once, that's why I need help with this. Essentially, it's a collision detection check in which the character has to turn around if they reach a vertical wall of at least 7 pixels in height, but  - all 7 pixels have to be filled. The following code works, but I want to try and simplify it if possible:

if HasPixelAt(X, Y) and HasPixelAt(X, Y -1)
  and HasPixelAt(X, Y -2) and HasPixelAt(X, Y -3)
  and HasPixelAt(X, Y -4) and HasPixelAt(X, Y -5)
  and HasPixelAt(X, Y -6) and HasPixelAt(X, Y -7) then
begin
...
end;

 

Thanks in advance, I hope this explains the question a little better.

I see no other way to do this for vertical walls, since the memory locations of the pixels in question are not contiguous if the storage used has the typical bitmap layout.

 

But you may be approaching the whole problem the wrong way. Instead of working directly with the bitmap image of the rendered scene you should build the scene as a list of objects, each of which represents an element of the scene. In this case the vertical wall would be one such object, storing its location and dimensions internally. You then walk over the list of objects and test whether the point (X,Y) falls inside such an object. This search can be sped up considerably by storing the objects in a tree or list sorted on coordinates. I have never worked in this area myself, but it is a common problem in game design, there should be plenty of literature around on the subject.

[programmerdelphi2k 237]

maybe when you run your code in "quantICUM" cpu you can have it!  8 states in 1... for now, just 2 in 1

Edited May 7, 2023 by programmerdelphi2k

[dummzeuch 1505]

37 minutes ago, Willicious said:

I hope this explains the question a little better.

See my first answer above. It contains the code you need, just replace (x, y-i) with HasPixelAt(x, y-i).

 

https://en.delphipraxis.net/topic/8900-for-i-x-to-y-inclusive-how/?do=findComment&comment=74894

[Willicious 8]

Edited

Edited May 7, 2023 by Willicious

[Willicious 8]

23 minutes ago, dummzeuch said:

See my first answer above. It contains the code you need, just replace (x, y-i) with HasPixelAt(x, y-i).

 

https://en.delphipraxis.net/topic/8900-for-i-x-to-y-inclusive-how/?do=findComment&comment=74894

Hmm, I can't get this to work. What is "Res :=" in this example?

Edited May 7, 2023 by Willicious

[programmerdelphi2k 237]

@Willicious

using your analogy of one set within another, and not the whole set, seems to me similar to what Delphi uses when adding or removing "an item" in an options property, i.e.:
AOption := AOption + [ItemX] or AOption := AOption - [ItemX]  (or using multiplication, etc...)
However, a "Set" is restricted to 256 elements, and, to circumvent this limit, you would have to use a List or Array to store the values, and later compare whether or not they are contained in the container!

 

other hazards could occur in an unattended task...

implementation

{$R *.dfm}

type
  TMyEnums     = (xx1, xx2, xx3, xx4, xx5, xx6, xx7, xx8, xx9, xx10);
  TMyEnumsSet = set of TMyEnums;

procedure TForm1.Button1Click(Sender: TObject);
var
  a: TMyEnumsSet;
  b: TMyEnumsSet;
  c: TMyEnumsSet;
  d: TMyEnums;
  e: TMyEnums;
begin
{$R-}
  // Intersection of Sets
  d := TMyEnums(11); // out of range, then... {$R-}...{$R+}
  e := TMyEnums(0);
  //
  a := [xx1, xx2, xx3, xx4, xx5, xx6, xx7, xx8, xx9, xx10];
  b := [xx2, xx3, d, xx1, xx8, xx2, e];
  c := a * b; // = xx1, xx2, xx3, xx8
{$R+}
end;

end.

 

maybe this post can help to add /remove new elements in Set

https://stackoverflow.com/questions/33173534/how-do-i-check-or-change-which-set-elements-are-present-using-rtti

Edited May 7, 2023 by programmerdelphi2k

[dummzeuch 1505]

1 hour ago, Willicious said:

Hmm, I can't get this to work. What is "Res :=" in this example?

Res is a Boolean variable.