What does this syntax mean? 2

[DaveC426913]

What does this syntax mean:

function foo($bar=0) {
...
}

I don't understand the parameter. Is it equivalent to an assignment? as in:

function foo() {
$bar=0;
...
}

or perhaps, an optional param?:

function foo($bar) {
if (is_null($bar)){
bar = 0;
}
...
}


Here it is in context. The actual variable is $shift (in case that's a reserved variable or something):


$newCal->displayMonth($mes);

...

function displayMonth($shift=0) {
...
$firstday_month_ts = mktime(0,0,0,date("n")+$shift,1,date("Y")); // first day of the month
$lastday_month_ts = mktime(0,0,0,date("n")+$shift+1,0,date("Y")); // last day of the month
...
if ($shift==0 && $today_ts==$day_i_ts) {
}

#$mes would be an integer representing a month

 

[danomac]

If you call the function without specifying the parameter, in that example it is set to 0. In that case the parameter is not required to call the function.

 

[vacunita]

That means that if no variable is passed to the function the default value of bar will be 0.

if you do this:

function foo($bar=0){
echo $bar;
}


and thne call foo:

foo();
$var="Somevalue";
foo($var);

this will output:

0
Somevalue

----------------------------------
Ignorance is not necessarily Bliss, case in point:
Unknown has caused an Unknown Error on Unknown and must be shutdown to prevent damage to Unknown.

 

[DaveC426913]

OK thanks. Then there must be some other reason why I'm seeing "" instead of 0.

 

[ingresman]

Post the code and some test results then, I'm sure someone will spot it.

 

[DaveC426913]

Actually, when I added this bit:

if ($shift==0){ echo "True"; }

I got "True".

 

[theEclipse]

I have noticed that sometimes when you print or echo variables that have the integer value of zero (or a boolean false) it doesnt print anything. Its kind of annoying if you ask me.

If I come across something that looks as if it might be doing something like that I usually turn the print statement into something like this:

print "|" . $shift . "|!" . (!$shift) . "|";

that way it will print out something no matter what.

 

[DaveC426913]

Yes. I truncated it. It's actually:

if ($shift==0){ echo "True"; }
else{ echo "False"; }

The other thing I do is:

echo "[".$i."]";

That way, I can be sure I'm not just missing it.