transient lazy vals in constructor
I would like to declare a @transient lazy val as a constructor parameter. Example:
class Foo(@transient lazy val foo: FooFoo).
Scala 2.11 does not yet seem to have support for this. The suggested way in the issue linked above is to declare another lazy val as a class member, and have it point to the call-by-name constructor parameter.
Example (from the issue):
def f[a](lazy t: a) =
..
should be equivalent to:
def f[a](t': => a) =
lazy val t: a = t'
..
My question is: Will the second method ensure that the lazy val is correctly initialized upon deserialization on being referenced for the first time?
scala lazy-evaluation
asked Nov 14 '18 at 0:35
lostsoul29lostsoul29
3451515
add a comment |
I would like to declare a @transient lazy val as a constructor parameter. Example:
class Foo(@transient lazy val foo: FooFoo).
Scala 2.11 does not yet seem to have support for this. The suggested way in the issue linked above is to declare another lazy val as a class member, and have it point to the call-by-name constructor parameter.
Example (from the issue):
def f[a](lazy t: a) =
..
should be equivalent to:
def f[a](t': => a) =
lazy val t: a = t'
..
My question is: Will the second method ensure that the lazy val is correctly initialized upon deserialization on being referenced for the first time?
scala lazy-evaluation
asked Nov 14 '18 at 0:35
lostsoul29lostsoul29
3451515
add a comment |
I would like to declare a @transient lazy val as a constructor parameter. Example:
class Foo(@transient lazy val foo: FooFoo).
Scala 2.11 does not yet seem to have support for this. The suggested way in the issue linked above is to declare another lazy val as a class member, and have it point to the call-by-name constructor parameter.
Example (from the issue):
def f[a](lazy t: a) =
..
should be equivalent to:
def f[a](t': => a) =
lazy val t: a = t'
..
My question is: Will the second method ensure that the lazy val is correctly initialized upon deserialization on being referenced for the first time?
scala lazy-evaluation
asked Nov 14 '18 at 0:35
lostsoul29lostsoul29
3451515
I would like to declare a @transient lazy val as a constructor parameter. Example:
class Foo(@transient lazy val foo: FooFoo).
Scala 2.11 does not yet seem to have support for this. The suggested way in the issue linked above is to declare another lazy val as a class member, and have it point to the call-by-name constructor parameter.
Example (from the issue):
def f[a](lazy t: a) =
..
should be equivalent to:
def f[a](t': => a) =
lazy val t: a = t'
..
My question is: Will the second method ensure that the lazy val is correctly initialized upon deserialization on being referenced for the first time?
scala lazy-evaluation
scala lazy-evaluation
asked Nov 14 '18 at 0:35
lostsoul29lostsoul29
3451515
asked Nov 14 '18 at 0:35
lostsoul29lostsoul29
3451515
edited Nov 14 '18 at 0:42
lostsoul29
asked Nov 14 '18 at 0:35
lostsoul29lostsoul29
3451515
asked Nov 14 '18 at 0:35
lostsoul29lostsoul29
3451515
asked Nov 14 '18 at 0:35
lostsoul29lostsoul29
3451515
3451515
add a comment |
add a comment |
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