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Answer by Mike 3
the function that you call needs to be of type IEnumerator - this is done implicitly when you use yield in the function, but if you add in a return (or it is possible to get through the function...
View ArticleAnswer by Absyss
what Mike meant is following:function myFirstFunction():IEnumerator { ... } function mySecondFunction():IEnumerator { ... } ... yield myFirstFunction(); yield mySecondFunction(); this way the compiler...
View ArticleAnswer by Techsuport
you could do something like a call back, so even though you cant get the return right away, you can have function B call function D in the same class as function A to continue on where A left off....
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