PHP function flags, how?
Asked Answered
R

1

21

I'm attempting to create a function with flags as its arguments but the output is always different with what's expected :

define("FLAG_A", 1);  
define("FLAG_B", 4);  
define("FLAG_C", 7);  
function test_flags($flags) {  
 if($flags & FLAG_A) echo "A";  
 if($flags & FLAG_B) echo "B";  
 if($flags & FLAG_C) echo "C";   
}  
test_flags(FLAG_B | FLAG_C); # Output is always ABC, not BC  

How can I fix this problem?

Repetitious answered 28/8, 2010 at 5:16 Comment(0)
O
37

Flags must be powers of 2 in order to bitwise-or together properly.

define("FLAG_A", 0x1);
define("FLAG_B", 0x2);
define("FLAG_C", 0x4);
function test_flags($flags) {
  if ($flags & FLAG_A) echo "A";
  if ($flags & FLAG_B) echo "B";
  if ($flags & FLAG_C) echo "C";
}
test_flags(FLAG_B | FLAG_C); # Now the output will be BC

Using hexadecimal notation for the constant values makes no difference to the behavior of the program, but is one idiomatic way of emphasizing to programmers that the values compose a bit field. Another would be to use shifts: 1<<0, 1<<1, 1<<2, &c.

Overbold answered 28/8, 2010 at 5:18 Comment(1)
Dude, the output expectations from Your comment isn't right. Because the output will always be otherwise - BC rather than ABC and this is a right behaviour for the program.Fresh

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