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  • \$\begingroup\$ Wow, TIL about the ?.() operator \$\endgroup\$ Commented Jan 6, 2022 at 16:53
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    \$\begingroup\$ @pxeger The () isn't really part of the operation. .? is the 'optional chaining' operator. It's mostly used with variables - e.g. a?.b?.c will result in c if both a and b are defined, and will result in undefined if either a or b is undefined/null. But it can also be used with indexing (a?.[i]) and functions (f?.(args)), of this this answer uses the second with map?.(f). :) \$\endgroup\$ Commented Jan 6, 2022 at 17:19
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    \$\begingroup\$ @KevinCruijssen That's not right; a.?b.?c will result in a.b.c if a and a.b are defined, unrelated to the variables b and c (that would be ??). ?.() is special-case syntax AFAICT \$\endgroup\$ Commented Jan 6, 2022 at 17:30
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    \$\begingroup\$ @pxeger a.?b.?c isn't valid JavaScript. I think you meant a?.b?.c? Partially my fault for that .? instead of ?. typo in the first line of my comment.. To correct & clarify part of my previous comment: ?. is the 'optional chaining' operator. It's mostly used with variables - e.g. a?.b?.c will result in a.b.c if both a and a.b are defined, and will result in undefined if either a or a.b is undefined/null. \$\endgroup\$ Commented Jan 6, 2022 at 17:54
  • \$\begingroup\$ @pxeger Anyway, it's still all the ?. operator, as can be seen in the link I send in my first comment. Whether you use it as obj?.val, arr?.[index] or func?.(args) doesn't matter too much. \$\endgroup\$ Commented Jan 6, 2022 at 17:56