A neater way of doing this?

SteveZX81

I have this line

If (ti[0] == 3) and (ti[1] == 3) and (ti[2] == 3) and (ti[3] == 3) and (ti[4] == 3) and (ti[5] == 3) and (ti[6] == 3) and (ti[7] == 3) and (ti[8] == 3) and (ti[9] == 3)

I wish to find a much neater/shorter way of doing it, what do you suggest?
(I only wish something to trigger if all 10 of ti[0-9] are equal to 3. not one or two, but all of them)

Jaywalker

allTrue = true
for i = 0 to 10 loop
if ti[i] != 3 then
allTrue = false
break
endif
repeat

if allTrue then
 //Do what you gotta do

Discostew

Had something similar to what @Jaywalker posted, but mine went through all of them rather than breaking the moment it found something not true. If this is something you plan to do for multiple different instances, you could convert it to a function and just call the function each time instead

function isAllEqual( arr, tst )
  ret = true
  for i = 0 to len( myarray ) loop
    if( arr[ i ] != tst ) then
      ret = false
      break
    endIf
  repeat
return ret

if ( isAllEqual( ti , 3 ) ) then
  // doot doot doot
else
  // nah nah nah
endIf

SteveZX81

Thank you! it works.

ITzTravelInTime

Remember that those kinds or ripetitve checks in which for example just one index changes beetween each condition can be always be done using a for cycle

donaldp

On line 4, i read != what is the difference between == and !=

I am just at the beginning of reading the help section

Thanks

Discostew

@donaldp said in A neater way of doing this?:

On line 4, i read != what is the difference between == and !=

I am just at the beginning of reading the help section

Thanks

== means "is equal to"
!= means "is not equal to"

donaldp

@Discostew thanks!