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Consider the following Erlang function: mystery( 0 ) -> 2; mystery( 1 ) -> 3; mystery( N ) when N > 1 -> mystery( N
Consider the following Erlang function:
mystery( 0 ) -> 2;
mystery( 1 ) -> 3;
mystery( N ) when N > 1 -> mystery( N 1 ) * mystery( N 2 ).
Show what would be returned by the following function calls. If the call would never finish, just put Never finish.
- mystery( 1 )
- mystery( 2 )
- mystery( 4 )
- mystery( 6 )
- mystery( -1 )
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