or || not working

jonSS

hello, anyone know why this doesnt work ?

var aaa1 = 0;


	if ( aaa1 != 0 || aaa1 != 1 ):
		print(aaa1);

	match state:
		"st_idle":
			aaa1 = 0;
		"st_walk":
			aaa1 = 1;

log----

! --- Debugging process started --- ! Godot Engine v3.4.stable.mono.official.206ba70f4 - https://godotengine.org !
! 0 ! 0 ! 0 ! 1 ! 0 ! 0 ! 1 ! 0 ! 0 ! 1 ! 0 ! 0 ! 1 ! 0 ! 1 ! 0 ! --- Debugging process stopped ---

and this is working? :#
I though the && would check if aaa1 would not equal 0 and would also not equal 1 ?

	if ( aaa1 != 0 && aaa1 != 1 ):
		print(aaa1);

log----

! --- Debugging process started --- ! Godot Engine v3.4.stable.mono.official.206ba70f4 - https://godotengine.org ! OpenGL ES Batching: ON !
! --- Debugging process stopped ---

another example:

if ( state != "st_idle" || state != "st_walk" ):
	print(state);

	match state:
		"st_idle":
			aaa1 = 0;
		"st_walk":
			aaa1 = 1;

log--

! --- Debugging process started --- ! Godot Engine v3.4.stable.mono.official.206ba70f4 - https://godotengine.org ! OpenGL ES Batching: ON ! st_idle ! st_idle ! st_idle ! st_walk ! st_walk ! st_walk ! st_walk ! st_idle ! st_walk ! st_walk ! st_idle ! st_walk ! st_idle ! st_walk ! st_idle ! st_walk ! st_idle ! st_walk ! st_idle ! st_walk ! st_idle ! st_walk ! st_idle ! st_walk ! st_idle ! st_walk ! st_idle ! st_walk ! --- Debugging process stopped ---

spacecloud

You're checking if aaa1 is not 0 or it's not 1, so the expression will only be false if aaa1 is 0 and 1 at the same time.

jonSS

@spacecloud said: You're checking if aaa1 is not 0 or it's not 1, so the expression will only be false if aaa1 is 0 and 1 at the same time.

weither i got confused with something, or i need some sleep

what about now ? aaa1 is never 0 or 1

var aaa1 = "left";

	if ( aaa1 != "right" || aaa1 != "left" ):
		print(aaa1);

	match state:
		"st_idle":
			aaa1 = "left";
		"st_walk":
			aaa1 = "right";

print log:

! --- Debugging process started --- ! Godot Engine v3.4.stable.mono.official.206ba70f4 - https://godotengine.org ! OpenGL ES Batching: ON ! left ! left ! left ! left ! left ! left ! left ! left ! left ! left ! left ! right ! right ! right ! right ! right ! right ! right ! left ! left ! left ! left ! left ! left ! left ! left ! left ! left ! left ! right ! right ! right ! right ! right ! right ! right ! right ! left ! left ! left ! left ! left ! left ! left ! left ! left ! left ! --- Debugging process stopped ---

spacecloud

Same, it will only be false when aaa1 is "right" and "left".

Basically,

if (a != some_value || a != some_other_value)

is the same as

if (a == some_value && a == some_other_value)

jonSS

@spacecloud said: Same, it will only be false when aaa1 is "right" and "left".

Basically,
if (a != some_value || a != some_other_value)
is the same as
if (a == some_value && a == some_other_value)

thanks, i think i understand now. Iam more used to check for states this way: if ( state == "st_idle" || state == "st_walk" ): print(state);

---------------//---------//---------------- in doing this:

        	if ( state != "st_idle" || state != "st_walk" ):
        		print(state);

i was doing the same thing has this:

	if ( state != "st_idle" ):
		print(state);
	elif ( state != "st_walk" ):
		print(state);

//-----------//-------------

    	if ( state != "st_idle" ):
    		print(state);
    	if ( state != "st_walk" ):
    		print(state);