Yes, you're correct! Originally omitted exists when not testing for the final value--then overgeneralized. However, testing $Hash{$neededVal}{param1}{key1} would fail when $Hash{$neededVal}{param1}{key1} == 0, but expliciting testing the value, as you've shown, would remedy that.
In reply to Re^3: Checking if hash value exists without implicitly defining it
by Kenosis
in thread Checking if hash value exists without implicitly defining it
by Doctrin
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