PHP function flags, how?

Question

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?

Ale · Accepted Answer · 2015-06-15 00:04:22Z

38

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.

edited Jun 15, 2015 at 0:04

Ale

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answered Aug 28, 2010 at 5:18

Ignacio Vazquez-Abrams

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1 Comment

Arthur Kushman Over a year ago

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.

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