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That's a source operation. You aren't putting values into the collection, you are examining values in the collection.
by TheCoelacanth 8y ago
That's a source operation. You aren't putting values into the collection, you are examining values in the collection.
- munificent 8y agoYes, except the type argument appears in parameter position, which is typically contravariant.
- TheCoelacanth 8y agoOkay, but it's not. Make source vs. sink explicit and it will be obvious because the method will be defined on source and not on sink.