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    PHP find output programs (User-defined functions) | set 1
    
    
    
    
	    Find the output of PHP programs | User-defined functions | Set 1: Enhance the knowledge of PHP User-defined functions concepts by solving and finding the output of some PHP programs.
	    
		    Submitted by Nidhi, on January 22, 2021
	    
    
    
    Question 1:
<?php
    function fun()
    {
        printf("Hello World");
    }
    echo ("Hello India");
?>
Output:
Hello India
    Explanation:
    In the above program, we defined a user-define function. But we did not call function fun() anywhere then it will not print "Hello World", only "Hello India" will print on the webpage.
    Question 2:
<?php
    fun();
    
    function fun()
    {
        printf("Hello World<br>");
    }
    
    echo ("Hello India");
?>
Output:
Hello World
Hello India
    Explanation:
    In the above program, we defined a user define function. Then we called function fun() then it will print "Hello World", and  "Hello India" will print on the webpage.
    Question 3:
<?php
    $A = 5;
    $B = 10;
    
    fun($A, $B);
    printf("%d, %d", $A, $B);
    
    function fun(int $A, int $B)
    {
        $A = 20;
        $B = 30;
    }
?>
Output:
5, 10
    Explanation:
    In the above program, we defined a user define function fun() that contains two parameters $A and $B. We modified the values of $A and $B, but it will not reflect at the calling position. Because here we used the call by value mechanism for parameter passing. Then "5,10" will be printed on the webpage.
    Question 4:
<?php
    $A = 5;
    $B = 10;
    
    fun(&$A, &$B);
    printf("%d, %d", $A, $B);
    
    function fun(int $A, int $B)
    {
        $A = 20;
        $B = 30;
    }
?>
Output:
PHP Parse error:  syntax error, unexpected '&' in 
/home/main.php on line 5
    Explanation:
    The above program will generate syntax error because we cannot use ampersand '&' operator in the function call. To implement call by reference mechanism for parameter passing. We need to use ampersand '&' operator in the function definition.
    Question 5:
<?php
    $A = 5;
    $B = 10;
    
    fun($A, $B);
    printf("%d, %d", $A, $B);
    
    function fun(int & $A, int & $B)
    {
        $A = 20;
        $B = 30;
    }
?>
Output:
20, 30
    Explanation:
    In the above program, we defined a user define function fun() that contains two parameters $A and $B. We modified the values of $A and $B, but it will also reflect at calling position. Because here we used the call by reference mechanism for parameter passing. Then "20,30" will be printed on the webpage.
    
    
    
    
    
    
    
  
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