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How to implement a bitmask in php?

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php

bitmask

I'm not sure if bitmask is the correct term. Let me explain:

In php, the error_reporting function can be called multiple ways:

// Report simple running errors error_reporting(E_ERROR | E_WARNING | E_PARSE);  // Reporting E_NOTICE can be good too (to report uninitialized // variables or catch variable name misspellings ...) error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);  // Report all errors except E_NOTICE // This is the default value set in php.ini error_reporting(E_ALL ^ E_NOTICE); 

I got the term bitmask from the php.net page here

Anyway the point of this is, I have implemented a SIMPLE method called ls which returns the contents of a directory.

This function takes 3 args... ( $include_hidden = false, $return_absolute = false, $ext = false )

So when i call the function, i set how i want the results. Whether i want the results to return hidden directories, whether i want basenames only etc.

so when i call the function i'm writing

ls(true, false, true) ls(false, false, true) ls(true, true, true) etc... 

I thought it would be much more readable if i could just flag how i want the data returned?

so something like:

ls( INCLUDE_HIDDEN | HIDE_EXTS ); ls( SHOW_ABSOLUTE_PATHS | HIDE_EXTS ); 

etc...

How would i implement this in terms of testing which flags have been called?

like image 365
AlexMorley-Finch Avatar asked Aug 09 '12 09:08

AlexMorley-Finch


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1 Answers

It's quite simple actually. First a bit of code to demonstrate how it can be implemented. If you don't understand anything about what this code is doing or how it works, feel free to ask additional questions in the comments:

const FLAG_1 = 0b0001; // 1 const FLAG_2 = 0b0010; // 2 const FLAG_3 = 0b0100; // 4 const FLAG_4 = 0b1000; // 8 // Can you see the pattern? ;-)  function show_flags ($flags) {   if ($flags & FLAG_1) {     echo "You passed flag 1!<br>\n";   }   if ($flags & FLAG_2) {     echo "You passed flag 2!<br>\n";   }   if ($flags & FLAG_3) {     echo "You passed flag 3!<br>\n";   }   if ($flags & FLAG_4) {     echo "You passed flag 4!<br>\n";   } }  show_flags(FLAG_1 | FLAG_3); 

Demo


Because the flags are integers, on a 32-bit platform you define up to 32 flags. On a 64-bit platform, it's 64. It is also possible to define the flags as strings, in which case the number of available flags is more or less infinite (within the bounds of system resources, of course). Here's how it works in binary (cut down to 8-bit integers for simplicity).

FLAG_1 Dec:    1 Binary: 00000001  FLAG_2 Dec:    2 Binary: 00000010  FLAG_3 Dec:    4 Binary: 00000100  // And so on... 

When you combine the flags to pass them to the function, you OR them together. Let's take a look at what happens when we pass FLAG_1 | FLAG_3

  00000001 | 00000100 = 00000101 

And when you want to see which flags were set, you AND the bitmask with the flag. So, lets take the result above and see if FLAG_3 was set:

  00000101 & 00000100 = 00000100 

...we get the value of the flag back, a non-zero integer - but if we see if FLAG_2 was set:

  00000101 & 00000010 = 00000000 

...we get zero. This means that you can simply evaluate the result of the AND operation as a boolean when checking if the value was passed.

like image 62
DaveRandom Avatar answered Sep 23 '22 21:09

DaveRandom