[Lazarus] How should "delete to end of word behave"

Marius fpclazarus at home.nl
Fri Nov 7 21:27:51 CET 2008


I'm a of the option to delete the spaces behind the word also to make 
the editor turbo and delphi compatible. Its not because i want it to be 
fully compatible, but handling it this way requires a lot less 
keystrokes if you want to delete the first 4 words in a sentence.

I was testing a bit more last night and i was wundering about the 
following situation:
a:= 'foo_bar| and xyz';
a:= 'foo_bar xyz';
Is this how it should behave ?
(watch the spaces)

Greetings,
Marius


Martin Friebe wrote:
> In 0.9.26 Lazarus deletes to the end of the current token. (Note current 
> SVN behaves different)
>  |foo_bar := 1;
>
> deletes as follows
>  foo_bar
>  <space>
>  :=
>  <space>
>  1
>  ;
>
> There is a request to include deletion of the space, into the deletion 
> of the word before it
>  foo_bar<space>
>  :=<space>
>  1
>  ;
>
> While this would be no problem if the cursor was at the beginning of a 
> word, this may be an issue if the cursor is in the middle of o word
>   foo_b|ar  and  xyz
> would be    foo_b|and  xyz
> instead of   foo_b|  and  xyz
>
> On the other hand, the 0.9.26 behaviour means that
>   foo_bar|  and  xyz
> means 2 ctrl-t, to delete "and", since the first only deletes the <space>
>
> see also http://bugs.freepascal.org/view.php?id=12583
>
> --------------
> My opinion is that handling the spaces as n entity of there own, which 
> can and will be deleted by a ctrl-t of its own, allows the best flexibility.
> Always deleting the spaces with the current word, leads to unwanted 
> situation, if the cursor is in the middle of a word.
>
> A 3rd option is to make it context sensitive, depending on the cursor 
> being at the start of the word (open-office and word do that). I am not 
> necessarily a fan of it.
>
>
> Any opinions?
>
>   




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