; syntax issue

jordi · July 15, 2017 at 2:18 AM

I just recognised that sometimes (calc) we have to use ; mandatory

like in this case, where that would work :

Code

set(myOtherVar, 10);
if(%1 !== null, calc(myVar, %1 + myOtherVar); );
debug('myVar value', myVar);

and that would not :

Code

set(myOtherVar, 10);
if(%1 !== null, calc(myVar, %1 + myOtherVar));
debug('myVar value', myVar);

With set and other actions you can just skip these "extra" semicolon, am I right ?

klaus.krpano · July 15, 2017 at 11:22 PM

Hi,

yes, when no action call is following the ';' at the end of the action is optional.

That means both of your code examples would work.

Best regards,
Klaus

jordi · July 27, 2017 at 1:26 PM

I know, but just copy and execute my code and you will see that not both of them work... in 1.19-pr0

klaus.krpano · August 2, 2017 at 9:33 PM

Hi,

that's different reason / special case here - calc(...) (without the ';') has a special meaning when used as parameter - please see here:
https://krpano.com/docu/actions/#calc

In this case (without ';'):

Code

if(%1 !== null, calc(myVar, %1 + myOtherVar));

it would be that situation - using calc() to resolve a parameter (the parameter of the if() action):

Code

action( calc(expression), ... )

and 'myVar, %1 + myOtherVar' would be an invalid expression - mainly because of the ',' character.

But in this case (with the ';'):

Code

if(%1 !== null, calc(myVar, %1 + myOtherVar););

it would be that situation - that means a normal action call:

Code

calc(variable, expression)

and therefore this case is working.

That means it works basically as intended but I agree that can be confusing ;-).

Best regards,
Klaus

; syntax issue

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