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Supertype calculations in alternatives and binary expressions drop {inf} precision information when mixing number{inf} with integer types (number{0} or int). #1004

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svott7 opened this issue Jul 10, 2024 · 0 comments

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svott7 commented Jul 10, 2024

image

The first two examples compute the type as expected. However, in the last two examples we get int and number{0} as result of the supertype calculation instead of number{inf}.

IMO the type calculation should preserve the {inf} precision property.

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