wright67uk Posted April 4, 2012 Share Posted April 4, 2012 Im trying to understand the code below. (taken from an O'REILLY book) Im trying to get my head around this, so can anyone tell me if im thinking about this in the right way; The function (fix_names) borrows the values from $a1,$a2 and $a3 and puts them into $n1,$n2, and $n3. It then processes the strings contained within these newly created variables and then returns the processed values back into $a1,$a2,and $a3? The variables $n1,$n2, and $n3 cannot be echoed as they were created within the function.? Please correct me if im wrong. <html> <head> </head> <body> <?php $a1 = "EDWARD"; $a2 = "thomas"; $a3 = "wriGHT"; fix_names($a1,$a2,$a3); echo $a1." ".$a2." ".$a3; function fix_names(&$n1,&$n2,&$n3) { $n1 = ucfirst(strtolower($n1)); $n2 = ucfirst(strtolower($n2)); $n3 = ucfirst(strtolower($n3)); } ?> </body> </html> Quote Link to comment Share on other sites More sharing options...
dragon_sa Posted April 4, 2012 Share Posted April 4, 2012 - You initially declare and set your variables $a1, $a2, $a3. - You then call the function fix_names declaring that they are the 3 variables you want processed. - The variables inside the function are $n1, $n2, $n3, these are only available inside the function, and are set to the values of the 3 variables you declared when you call this function in this case $a1, $a2, $a3. - The function applies the formatting to the 3 variables and then returns the values to $a1, $a2, $a3 repectively. - You can call the same function many times like this <?php $a1 = "EDWARD"; $a2 = "thomas"; $a3 = "wriGHT"; $b1 = "bOB"; $b2 = "FRed"; $b3 = "johN"; $dave = "daVe"; $mark = "MARK"; $frank = "jasOn"; fix_names($a1,$a2,$a3); echo $a1." ".$a2." ".$a3."<br/>"; fix_names($b1,$b2,$b3); echo $b1." ".$b2." ".$b3."<br/>"; fix_names($dave,$mark,$frank); echo $dave." ".$mark." ".$frank."<br/>"; function fix_names(&$n1,&$n2,&$n3) { $n1 = ucfirst(strtolower($n1)); $n2 = ucfirst(strtolower($n2)); $n3 = ucfirst(strtolower($n3)); } ?> Try it for yourself Quote Link to comment Share on other sites More sharing options...
mofm Posted April 4, 2012 Share Posted April 4, 2012 Ok to fully understand what's going on here you must understand that when you pass a variable by reference you are not sending the data contained within that variable but instead sending an address of memory where the data you want is located... this is very different then just sending data. When a function executes you must understand that it has a certain level of scope, ie it cant see/edit variables out side of its { } but if you pass a variable as a reference you allow the function to directly edit a piece of memory. Let me explain with a few examples Example 1 The basics of a function <?php $a=10; $b=addOne($a); // this is simple and you should understand this echo $a; //10 echo $b; //11 function addOne($var) { return $var ++; // This function cant directly edit $a as it is outside its scope so instead we must "return" the data from the variable } ?> Example 2 Using example 1 but instead passing $a as reference. <?php $a=10; addOne(&$a); // I pass $a as a reference using & echo $a; //11 function addOne($var) // now var doesn't contain 10 but instead it contians the location of $a { $var ++; // this is like doing $a++ but it goes through $var inorder to access $a } I hope this helps you get your head round this, its not that hard like anything in php once you get the concept Quote Link to comment Share on other sites More sharing options...
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