[GRiMMREAPER]

Quote:

Originally Posted by coool TIL This works PHP код: new arr[2] = {1,3}; if(arr[0 || 1])     print("Yea, it works.");  Got the above when lurking my old threads https://sampforum.blast.hk/showthread.php?tid=624112 People are saying that if(arr[1 || 2]) is invalid syntax... But today I checked it works.

For x || y, provided both x and y are numbers, will always return 1. You are then accessing arr[1], which indeed exists and is 3.

pawn Код:

new arr[2] = {1};

    if(arr[0 || 1]) {
        print("It works!");
    } else {
        print("It doesn't work."); // Expected output, because arr[0 || 1] returns true (1) and arr[1] doesn't exist.
    }