Default arguments

[nomadrw]

Hello everyone:

I new to this site and new to PHP.

I trying to learn PHP on my own and I don't understand this short example. I hoping someone can explain it to me.

Here is the code:

function tour_guild($city = "Gotham City",
                             $desc = "vast metropolis",
                             $how_many ="dozens",
                             $of_what = "costumed villains")
{
Print ("$city is a $desc filled with $how_many of $of_what. <br>");
}
tour_guild();
tour_guild("Chicago");
tour_guild("Chicago", wonderful city");
tour_guild("Chicago", "wonderful city",
                "teeming millions");
tour_guild("Chicago" "wonderful city",
                "gruff people with hearts of
                gold and hard_luck");

 

Not use how the lower parameter are printed (tour_guild) ie

tour_guild();

tour_guild("Chicago");

tour_guild("Chicago", wonderful city");

tour_guild("Chicago", "wonderful city",

                "teeming millions");

...

 

any help would be great.

 

 

 

 

[Alex]

Values for parameters given in the function declaration designate default values if that specific parameter happens to not be passed, if it is passed it'll just be overlooked. Example:

 

function doSomething($var = 50)
{
     echo $var;
}

/*
Because I don't pass the first parameter in this example it assumes the default value designated in 
the function declaration. ($var = 50). As a result this function will echo '50'.
*/
doSomething();

/*
This time I do supply the first parameter in the function call so the default value of 50 is 
overlooked, and it just assumes the value that I supply in the function call. As a result this time the function echos 'Hello'.
*/
doSomething('Hello');

 

[nomadrw]

Thanks AlexWD.

I get it now.

 

nomad