Can you explain this?

[applovr]

Hi! Glad there's another forum for PHP n00bs like me.

Can anyone explain what this function does?

function dict_check($d, $s) {
	if ($d["type"] != "dictionary")
		bark("not a dictionary");
	$a = explode(":", $s);
	$dd = $d["value"];
	$ret = array();
	foreach ($a as $k) {
		unset($t);
		if (preg_match('/^(.*)\((.*)\)$/', $k, $m)) {
			$k = $m[1];
			$t = $m[2];
		}
		if (!isset($dd[$k]))
			bark("dictionary is missing key(s)");
		if (isset($t)) {
			if ($dd[$k]["type"] != $t)
				bark("invalid entry in dictionary");
			$ret[] = $dd[$k]["value"];
		}

Thanks!

Alex

[rowanparker]

Well I can tell you what each line does.

Could you give an example of usage so we can see what would be passed to it?

[etully]

You pass it a dictionary ($d) and a list of words ($s). The words are in the form:

word:word:word(some-type):word:word:word(some-type)

It looks through all the words in $s and tells you when the dictionary doesn't have a definition for that word and when the parenthecized TYPE isn't the same as the dictionary's value for that word's type. Otherwise, it compiles a list of definitions of the words.

Of course, "dictionary", "word" and "type" are generic terms. It could just as easily be used to compile a list of motorcycle parts or to compile a list of politician's flaws.

[miramardesign]

<?php 
function dict_check($d, $s) {    //function declaration  taking 2 params d and s (descriptive vars make code easier to read!!)  

	if ($d["type"] != "dictionary")  // if the value of the array $d with the key type does not equal the string dictionary 
		bark("not a dictionary");  //call user defined function bark() and pass the string to it "not a dictionary"
	$a = explode(":", $s);  //break apart the variable $s into an array by using the : character and save it to $a;
	$dd = $d["value"];  //the var $dd equals the value of the array $d with the key "value"
	$ret = array();  // the var $ret is declared an array
	foreach ($a as $k) {  //loop through the array $a replacing its value w/ $k
		unset($t);   // make the var $t empty or null 
		if (preg_match('/^(.*)\((.*)\)$/', $k, $m)) {   // if $k matches $m via a  perl compatible regular expression  

			$k = $m[1];  // then the var $k equals the second value of the array $m
			$t = $m[2];//and the var $t equals the third value of the array $m
		}  // end if 
		if (!isset($dd[$k])) //if the value of the array $dd w/ the associative index of $k is not set 
			bark("dictionary is missing key(s)");  //call the function bark w/ the param 'dictionary is missing key..'
		if (isset($t)) {  // if the value of $t is set
			if ($dd[$k]["type"] != $t) // if the value of the array $dd with the associative key of the array $k with the associative key of 'type'
				bark("invalid entry in dictionary"); //call bark with string params
			$ret[] = $dd[$k]["value"];  //the array $ret equals the array $dd with the associative key of $k and the associative key of 'value'
		}// end if 
?>

Whew I did my best to translate that into english for fun practice 🙂

I might have messed up badly I think my parsers off kilter...

Anyways that code makes little sense because the vars are named as letters which have no meaning. As w/ anything in programming descriptive and verbose names help you know what each line is talking about by giving you hints at the var name level .

Good Luck
MiramarDesign
http://miramardesign.com
I code therefore I am.

[etully]

that code makes little sense

If you were unclear about what the function does, you could read my post from yesterday.