here's a love file in case you don't understand: (press P to advance to the next level)

**Edit: nevermind, i was dumb**

i have a function that calculates the position of an enemy in tiles (32x32 squares). i have the starting position of the enemies stored in a table. for some reason, when love.graphics.draw tries to draw the enemy onto the screen, it thinks the first value of the table (which by the way is 3) is a bool value, and i cant figure out why. there isnt a single bool value within the entire level.

here's a love file in case you don't understand: (press P to advance to the next level)**Edit: nevermind, i was dumb**

here's a love file in case you don't understand: (press P to advance to the next level)

Last edited by fixylol on Wed Apr 13, 2016 5:57 pm, edited 1 time in total.

- DaedalusYoung
- Party member
**Posts:**402**Joined:**Sun Jul 14, 2013 8:04 pm

This: seems to return false.

Code: Select all

`v[3] == 2 and v[2] + 1 or v[3] == 4 and v[2] - 1`

it'shouldnt set the value to false though :/DaedalusYoung wrote:This:seems to return false.Code: Select all

`v[3] == 2 and v[2] + 1 or v[3] == 4 and v[2] - 1`

- TheOdyssey
- Citizen
**Posts:**52**Joined:**Mon Jul 13, 2015 4:29 pm**Location:**Turn Around...

That does generates a boolean
Is the exact same thing as

Code: Select all

` v[2] = v[3] == 2 `

Code: Select all

```
if v[3] == 2 then
v[2] = true
end
```

- zorg
- Party member
**Posts:**2993**Joined:**Thu Dec 13, 2012 2:55 pm**Location:**Absurdistan, Hungary-
**Contact:**

Probably some precedence issues;
...well, technically, you could only enclose the whole clause after the **or**, but this is a bit more straightforward.

Code: Select all

```
-- v[3] == 2 and v[2] + 1 or v[3] == 4 and v[2] - 1 should be
something = ((v[3] == 2) and (v[2] + 1)) or ((v[3] == 4) and (v[2] - 1))
```

Me and my stuff True Neutral Aspirant. Why, yes, i do indeed enjoy sarcastically correcting others when they make the most blatant of spelling mistakes. No bullying or trolling the innocent tho.

Adding those parens won't change precedence; "==" > "and" > "or" (see PIL 3.5) so "A == B and C or D == E and F" is equivalent to "((A == B) and C) or ((D == E) and F)" ... if A ~= B and D ~= E then the expression does evaluate to false, though (noted earlier).

Code: Select all

`x = v[3] == 2 and v[2] + 1 or v[3] == 4 and v[2] - 1`

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```
if v[3] == 2 then
x = v[2] + 1
elseif v[3] == 4 then
x = v[2] - 1
else
x = false
end
```

- DPlayer234
- Prole
**Posts:**4**Joined:**Tue Apr 12, 2016 3:36 pm-
**Contact:**

The solution is pretty simple.

This: (lines 225-228)
would set v[1] to false if neither "v[3] == 1" or "v[3] == 3" return true. (Same for v[2])

Why? In case no statement in such a "chain" is true, it will return false.

What I did to prevent the crash was simple:
This code will now make it return 1 if both "v[3] == 1" and "v[3] == 3" return true (the "or 1"-part does that).

This: (lines 225-228)

Code: Select all

```
for i,v in pairs(CurrentLevel.Enemies) do
v[1] = v[3] == 1 and v[1] + 1 or v[3] == 3 and v[1] - 1
v[2] = v[3] == 2 and v[2] + 1 or v[3] == 4 and v[2] - 1
end
```

Why? In case no statement in such a "chain" is true, it will return false.

What I did to prevent the crash was simple:

Code: Select all

```
for i,v in pairs(CurrentLevel.Enemies) do
v[1] = v[3] == 1 and v[1] + 1 or v[3] == 3 and v[1] - 1 or 1
v[2] = v[3] == 2 and v[2] + 1 or v[3] == 4 and v[2] - 1 or 1
end
```

- zorg
- Party member
**Posts:**2993**Joined:**Thu Dec 13, 2012 2:55 pm**Location:**Absurdistan, Hungary-
**Contact:**

You mean it'll return 1 if both return false; if both returned true, then it'd set the value to the first in the chain.DPlayer234 wrote:This code will now make it return 1 if both "v[3] == 1" and "v[3] == 3" return true (the "or 1"-part does that).

Me and my stuff True Neutral Aspirant. Why, yes, i do indeed enjoy sarcastically correcting others when they make the most blatant of spelling mistakes. No bullying or trolling the innocent tho.

- DPlayer234
- Prole
**Posts:**4**Joined:**Tue Apr 12, 2016 3:36 pm-
**Contact:**

Oh yeah, right. Of course. My bad.zorg wrote:You mean it'll return 1 if both return false; if both returned true, then it'd set the value to the first in the chain.DPlayer234 wrote:This code will now make it return 1 if both "v[3] == 1" and "v[3] == 3" return true (the "or 1"-part does that).

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