[SyS]

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.

It is syntactically correct but it won't wont give the result what you expect to get. || is logical OR operator it returns true (1) if one of the operands is true (not 0) and returns false (0) if both operands are 0. arr[0 || 1] is evaluated to arr[1] in this case.