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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
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.
The text was updated successfully, but these errors were encountered:
The first two examples compute the type as expected. However, in the last two examples we get
int
andnumber{0}
as result of the supertype calculation instead ofnumber{inf}
.IMO the type calculation should preserve the {inf} precision property.
The text was updated successfully, but these errors were encountered: