I have a conditional in my program that I just can't seem to get working right. I've rewrote it several times, but can't think of anything else to try.
Code:
:If sub(Str1,Z,1)="1 and L2(1=-44 or sub(Str1,Z,1)="2 and L2(1)=-44
That's giving me an invalid dim when it comes to it. I tried saying "if the first thing AND the second or the third" but it didn't work (but didn't give me an error). Parenthises didn't seem to work like in other languages.
Btw, this is my first time working with strings. I wouldn't be surprised if it's a string-related problem.
Methinks you should close parentheses and quotes before "and"s and "or"s.
calc84maniac wrote:
Methinks you should close parentheses and quotes before "and"s and "or"s.
Yep, that did it! Thanks. Damn optimizations. 
Oh, well. But is there a way to save a few bytes and only type the "l2(1)=-44" once?
Code:
:sub(Str1,Z,1
:If L2(1)=-44 and (Ans="1" or Ans="2
calc84maniac wrote:
Code:
:sub(Str1,Z,1
:If L2(1)=-44 and (Ans="1" or Ans="2
That's odd. I thought I tried that setup... I must be slipping today. Thanks, again!
You're welcome.
Here's some more optimization:
Code:
:sub(Str1,Z,1
:If -44=L2(1)(Ans="1" or Ans="2
calc84maniac wrote:
You're welcome.
Here's some more optimization:
Code:
:sub(Str1,Z,1
:If -44=L2(1)(Ans="1" or Ans="2
Code:
:sub(Str1,Z,1
:If (-44=L2(1))(Ans="1" or Ans="2
The only way -44 can equal L2(1)(Ans="1" or Ans="2") is if:
(Ans="1" or Ans="2") is true
and
L2(1)=-44
which is exactly what it did before, only smaller.
calc84maniac wrote:
The only way -44 can equal L2(1)(Ans="1" or Ans="2") is if:
(Ans="1" or Ans="2") is true
and
L2(1)=-44
which is exactly what it did before, only smaller.
No! Look at the way you have it: first you do Ans = "1" or Ans="2, agreed? Next, it does the implicit multiplication of that times L2(1). Finally, it compares that to -44!! It's all order of operations: comparison generally comes after all arithmetic operations, and ()( counts as implicit multiplication.
Try mine and see for yourself if it works.
calc84maniac wrote:
Try mine and see for yourself if it works.
Yes, it works in this particular case since, for instance, you're testing L2(1) for a nonzero value. It won't work in all situations.
I only meant this particular case anyway.
calc84maniac wrote:
I only meant this particular case anyway.
Fair enough. carry on, then.
I thought using AND anyway is faster than implicit multiplication.
something1990 wrote:
I thought using AND anyway is faster than implicit multiplication.
it is slower than using ' and ', but for whatever reason, people still proclaim the glory of implicit and via multiplication...
I suppose because it's easier to read during debugging. Or else people are trying to be 1337 by using a shortcut.
Another string related question. How do you replace a character out of the middle of a string? Does it involve the inString command (I'm away from home and don't have my manual)?
:"1"->sub(Str1,4,1 doesn't work...
Say Str1="12345679", and you want to add an "8" between the "7" and "9".
Code: :sub(str1,1,7)+"8"+sub(str1.8.1 ->str1
Basically take the first section up until immediately before the space where you want to add a character, add the character, and then add the rest of the original string IN THAT ORDER.
Kuro wrote:
Say Str1="12345679", and you want to add an "8" between the "7" and "9".
Code: :sub(str1,1,7)+"8"+sub(str1.8.1 ->str1
Basically take the first section up until immediately before the space where you want to add a character, add the character, and then add the rest of the original string IN THAT ORDER.
Cool! Thanks, Kuro!
It comes from long experience with strings.
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