Only emit .try_into()? when a type is specified
#397
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tarcieri
reviewed
Feb 8, 2022
The purpose of this transformation is for when the caller specifies `type = "..."`. In that case, we decode into the specified type and then convert into actual type. However, in cases where no type is specified, we have to decode into an unspecified type `T` and then convert into the final type; but the compiler (obviously) cannot determine the type `T`. Therefore, when no `type` is specified, we drop the `.try_into()?` and just attempt to decode into the final type directly. Signed-off-by: Nathaniel McCallum <nathaniel@profian.com>
tarcieri
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Feb 8, 2022
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The purpose of this transformation is for when the caller specifies
type = "...". In that case, we decode into the specified type and then convert intoactual type. However, in cases where no type is specified, we have to decode
into an unspecified type
Tand then convert into the final type; but thecompiler (obviously) cannot determine the type
T.Therefore, when no
typeis specified, we drop the.try_into()?and justattempt to decode into the final type directly.
Signed-off-by: Nathaniel McCallum nathaniel@profian.com