array elements into numbers

[bored_onion]

if i have a 3 element array of which each is a number
eg:
[1,2,3]

how do i add these numbers together?
(in this case i would get 6)
i tried

Code: [Select]

_number = _array select 0 + _array select 1 + _array select 2
but it returned another array

thanks

[Blanco]

I'm not an expert with arrays but try this :

Code: [Select]

_array = [1,2,3]
_result = 0

_i = 0

#L
_value = _array select _i
_result = _result + _value
_i = _i + 1
?_i < count _array : goto "L"

hint format ["DEBUG = %1",_result]


But there must be a much easier way...

[Mud_Spike]

Same concept as Blanco proposed but with less code:

Code: [Select]

_res = 0
"_res = _res + _x" foreach _array

Edit:
Just a note on your own example:

Code: [Select]

_number = _array select 0 + _array select 1 + _array select 2Because of operator precedence the result is not what you think it is, in such cases (or whenever you are unsure), use parens to group the individual expressions.:

Code: [Select]

_number = (_array select 0) + (_array select 1) + (_array select 2)

[General Barron]

Just use parenthesis, and it should work:

_number = (_array select 0) + (_array select 1) + (_array select 2 )

Its better to overuse parenthesis then to under use them. Lots of frustration has been caused by me not using enough parenthesis before...

[bored_onion]

cheers guys i appreciate the help

[dmakatra]

Just use parenthesis, and it should work:

_number = (_array select 0) + (_array select 1) + (_array select 2 )

Its better to overuse parenthesis then to under use them. Lots of frustration has been caused by me not using enough parenthesis before...

(_number = (((_array select 0)) + (((_array select 1))) + (_array select (2))))

 ;D

If the problem is solved then press the little solve button mate.

:beat: *Gets Shot* :beat:

[C0LDSt33L]

He won't be able to solve until Tuesday (I think) because he's just gone off on some mini holiday.