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Posted by Stephen thangaraj at 18:57
Function in php

What Is the Scope of a Variable Defined in a Function?
The scope of a local variable defined in a function is limited with that function. Once the function is ended, its local variables are also removed. So you can not access any local variable outside its defining function. Here is a PHP script on the scope of local variables in a function:
<?php
?>
function myPassword() {
  $password = "U8FIE8W0";
  print("Defined inside the function? ". isset($password)."\n");
}
myPassword();
print("Defined outside the function? ". isset($password)."\n");
?>


This script will print:
Defined inside the function? 1
Defined outside the function?


What Is the Scope of a Variable Defined outside a Function?
A variable defined outside any functions in main script body is called global variable. However, a global variable is not really accessible globally any in the script. The scope of global variable is limited to all statements outside any functions. So you can not access any global variables inside a function. Here is a PHP script on the scope of global variables:
<?php
?>
$login = "pickzycenter";
function myLogin() {
  print("Defined inside the function? ". isset($login)."\n");
}
myLogin();
print("Defined outside the function? ". isset($login)."\n");
?>
This script will print:
Defined inside the function?
Defined outside the function? 1


How To Access a Global Variable inside a Function?
By default, global variables are not accessible inside a function. However, you can make them accessible by declare them as "global" inside a function. Here is a PHP script on declaring "global" variables:
<?php
?>
$intRate = 5.5;
function myAccount() {
  global $intRate;
  print("Defined inside the function? ". isset($intRate)."\n");
}
myAccount();
print("Defined outside the function? ". isset($intRate)."\n");
?>
This script will print:
Defined inside the function? 1
Defined outside the function? 1


How Values Are Returned from Functions?
If a value is returned from a function, it is returned by value, not by reference. That means that a copy of the value is return. Here is a PHP script on how values are returned from a function:
<?php
$favor = "vbulletin";
function getFavor() {
  global $favor;
  return $favor;
}
$myFavor = getFavor();
print("Favorite tool: $myFavor\n");
$favor = "phpbb";
print("Favorite tool: $myFavor\n");
?>
This script will print:
Favorite tool: vbulletin
Favorite tool: vbulletin
As you can see, changing the value in $favor does not affect $myFavor. This proves that the function returns a new copy of $favor.


How To Return a Reference from a Function?
To return a reference from a function, you need to:
  • Add the reference operator "&" when defining the function.
  • Add the reference operator "&" when invoking the function.
Here is a PHP script on how to return a reference from a function:
<?php
$favor = "vbulletin";
function &getFavorRef() {
  global $favor;
  return $favor;
}
$myFavor = &getFavorRef();
print("Favorite tool: $myFavor\n");
$favor = "phpbb";
print("Favorite tool: $myFavor\n");
?>
This script will print:
Favorite tool: vbulletin
Favorite tool: phpbb

As you can see, changing the value in $favor does affect $myFavor, because $myFavor is a reference to $favor.


How To Specify Argument Default Values?
If you want to allow the caller to skip an argument when calling a function, you can define the argument with a default value when defining the function. Adding a default value to an argument can be done like this "function name($arg=expression){}. Here is a PHP script on how to specify default values to arguments:
<?php
function printKey($key="download") {
  print("PHP $key\n");
}
printKey();
printKey("hosting");
print("\n");
?>
This script will print:
PHP download
PHP hosting


How To Define a Function with Any Number of Arguments?
If you want to define a function with any number of arguments, you need to:
  • Declare the function with no argument.
  • Call func_num_args() in the function to get the number of the arguments.
  • Call func_get_args() in the function to get all the arguments in an array.


Here is a PHP script on how to handle any number of arguments:
<?php
function myAverage() {
  $count = func_num_args();
  $args = func_get_args();
  $sum = array_sum($args);
  return $sum/$count;  
}
$average = myAverage(102, 121, 105);
print("Average 1: $average\n");
$average = myAverage(102, 121, 105, 99, 101, 110, 116, 101, 114);
print("Average 2: $average\n");
?>
This script will print:
Average 1: 109.33333333333
Average 2: 107.66666666667


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