[JuzDoiT]

warning 202: number of arguments does not match definition

problem is on this line

PHP код:

 "if (IsPlayerInRangeOfActor(playerid, LSAMMU_ACTOR, SFAMMU_ACTOR, LVAMMU_ACTOR))" //cuz i added LVAMMU_ACTOR 


PHP код:

stock InRangeOfAmmuActors(playerid)
{
    if (IsPlayerInRangeOfActor(playerid, LSAMMU_ACTOR, SFAMMU_ACTOR, LVAMMU_ACTOR))
    {
        return 1;
    }
    return 0;
} 


when i try without LVAMMU_ACTOR it work good

PHP код:

stock InRangeOfAmmuActors(playerid)
{
    if (IsPlayerInRangeOfActor(playerid, LSAMMU_ACTOR, SFAMMU_ACTOR))
    {
        return 1;
    }
    return 0;
} 


here is IsPlayerInRangeOfActor

Quote:

stock IsPlayerInRangeOfActor(playerid, actorid, Float:radius = 3.0) { new Float, Float:y, Float:z; if (GetActorPos(actorid, x, y, z)) { return IsPlayerInRangeOfPoint(playerid, radius, x, y, z); } return 0; }

[Toroi]

The warning is pretty straightforward, is not it?

Quote:

number of arguments does not match definition

The arguments you must to pass the function is what the function's header tells you to:

PHP код:

IsPlayerInRangeOfActor(playerid, actorid, Float:radius = 3.0) 


If what you want to do is to check if the player is near one or more actors, add an 'or' ( || ) separator in that conditional statement and add as many IsPlayerInRangeOfActor functions as you want, complying with the required arguments.

I can't give you an example because the function itself is pretty straightforward, as i said before.

[JuzDoiT]

nvm , fixed , thanks for the info