Having a companion object seems to maask implicit calls to apply?
Consider this code, which works:
case class Foo(s: String)
Some("bar").map(Foo)
It's technically calling Foo.apply, and that's implied.
But now if I have a companion object:
case class Foo(s: String)
object Foo
def apply(): Foo = Foo("default")
Some("bar").map(Foo)
Now this doesn't work because it is thinking of Foo.type, as in the object Foo. You have to explicitly say .map(Foo.apply)
Is this justScalaThings, or am I doing something wrong?
(Yes I know in this trivial example I could just default the parameter in the case class declaration, this is just an example. Lots of reasons you might declare a companion object.)
scala
asked Nov 15 '18 at 14:59
user435779user435779
212214
add a comment |
Consider this code, which works:
case class Foo(s: String)
Some("bar").map(Foo)
It's technically calling Foo.apply, and that's implied.
But now if I have a companion object:
case class Foo(s: String)
object Foo
def apply(): Foo = Foo("default")
Some("bar").map(Foo)
Now this doesn't work because it is thinking of Foo.type, as in the object Foo. You have to explicitly say .map(Foo.apply)
Is this justScalaThings, or am I doing something wrong?
(Yes I know in this trivial example I could just default the parameter in the case class declaration, this is just an example. Lots of reasons you might declare a companion object.)
scala
asked Nov 15 '18 at 14:59
user435779user435779
212214
add a comment |
Consider this code, which works:
case class Foo(s: String)
Some("bar").map(Foo)
It's technically calling Foo.apply, and that's implied.
But now if I have a companion object:
case class Foo(s: String)
object Foo
def apply(): Foo = Foo("default")
Some("bar").map(Foo)
Now this doesn't work because it is thinking of Foo.type, as in the object Foo. You have to explicitly say .map(Foo.apply)
Is this justScalaThings, or am I doing something wrong?
(Yes I know in this trivial example I could just default the parameter in the case class declaration, this is just an example. Lots of reasons you might declare a companion object.)
scala
asked Nov 15 '18 at 14:59
user435779user435779
212214
Consider this code, which works:
case class Foo(s: String)
Some("bar").map(Foo)
It's technically calling Foo.apply, and that's implied.
But now if I have a companion object:
case class Foo(s: String)
object Foo
def apply(): Foo = Foo("default")
Some("bar").map(Foo)
Now this doesn't work because it is thinking of Foo.type, as in the object Foo. You have to explicitly say .map(Foo.apply)
Is this justScalaThings, or am I doing something wrong?
(Yes I know in this trivial example I could just default the parameter in the case class declaration, this is just an example. Lots of reasons you might declare a companion object.)
scala
scala
asked Nov 15 '18 at 14:59
user435779user435779
212214
asked Nov 15 '18 at 14:59
user435779user435779
212214
asked Nov 15 '18 at 14:59
user435779user435779
212214
asked Nov 15 '18 at 14:59
user435779user435779
212214
asked Nov 15 '18 at 14:59
user435779user435779
212214
212214
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
In your first example, there is only Foo.apply fitting .map(Foo).
In the second example you have declared an object called Foo, superceding Foo.apply, which you then try to pass into the map. Because map expects a parameter of type String => A compilation fails.
So as far as I am concerned you have to use .map(Foo.apply).
This for example won't compile either:
case class Foo()
def map[A](f: Unit => A) = ???
map(Foo)
Try it out!
answered Nov 15 '18 at 15:13
Markus AppelMarkus Appel
942220
Thanks. After messing with it, that was also my conclusion, that since it's now ambiguous whether you mean Foo.apply or Foo.type, you have to just specify. Wanted to make sure I wasn't missing something. Thanks!
– user435779
Nov 15 '18 at 16:27
It's notFoo.type, it'sFoo. You are attempting to pass theobject Foo. You can't pass a type as a parameter (types are compile time only), those are two different things. Anyways, I am happy I could help.
– Markus Appel
Nov 15 '18 at 16:37
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
In your first example, there is only Foo.apply fitting .map(Foo).
In the second example you have declared an object called Foo, superceding Foo.apply, which you then try to pass into the map. Because map expects a parameter of type String => A compilation fails.
So as far as I am concerned you have to use .map(Foo.apply).
This for example won't compile either:
case class Foo()
def map[A](f: Unit => A) = ???
map(Foo)
Try it out!
answered Nov 15 '18 at 15:13
Markus AppelMarkus Appel
942220
Thanks. After messing with it, that was also my conclusion, that since it's now ambiguous whether you mean Foo.apply or Foo.type, you have to just specify. Wanted to make sure I wasn't missing something. Thanks!
– user435779
Nov 15 '18 at 16:27
It's notFoo.type, it'sFoo. You are attempting to pass theobject Foo. You can't pass a type as a parameter (types are compile time only), those are two different things. Anyways, I am happy I could help.
– Markus Appel
Nov 15 '18 at 16:37
add a comment |
In your first example, there is only Foo.apply fitting .map(Foo).
In the second example you have declared an object called Foo, superceding Foo.apply, which you then try to pass into the map. Because map expects a parameter of type String => A compilation fails.
So as far as I am concerned you have to use .map(Foo.apply).
This for example won't compile either:
case class Foo()
def map[A](f: Unit => A) = ???
map(Foo)
Try it out!
answered Nov 15 '18 at 15:13
Markus AppelMarkus Appel
942220
Thanks. After messing with it, that was also my conclusion, that since it's now ambiguous whether you mean Foo.apply or Foo.type, you have to just specify. Wanted to make sure I wasn't missing something. Thanks!
– user435779
Nov 15 '18 at 16:27
It's notFoo.type, it'sFoo. You are attempting to pass theobject Foo. You can't pass a type as a parameter (types are compile time only), those are two different things. Anyways, I am happy I could help.
– Markus Appel
Nov 15 '18 at 16:37
add a comment |
In your first example, there is only Foo.apply fitting .map(Foo).
In the second example you have declared an object called Foo, superceding Foo.apply, which you then try to pass into the map. Because map expects a parameter of type String => A compilation fails.
So as far as I am concerned you have to use .map(Foo.apply).
This for example won't compile either:
case class Foo()
def map[A](f: Unit => A) = ???
map(Foo)
Try it out!
answered Nov 15 '18 at 15:13
Markus AppelMarkus Appel
942220
In your first example, there is only Foo.apply fitting .map(Foo).
In the second example you have declared an object called Foo, superceding Foo.apply, which you then try to pass into the map. Because map expects a parameter of type String => A compilation fails.
So as far as I am concerned you have to use .map(Foo.apply).
This for example won't compile either:
case class Foo()
def map[A](f: Unit => A) = ???
map(Foo)
Try it out!
answered Nov 15 '18 at 15:13
Markus AppelMarkus Appel
942220
edited Nov 15 '18 at 15:22
answered Nov 15 '18 at 15:13
Markus AppelMarkus Appel
942220
answered Nov 15 '18 at 15:13
Markus AppelMarkus Appel
942220
answered Nov 15 '18 at 15:13
Markus AppelMarkus Appel
942220
942220
Thanks. After messing with it, that was also my conclusion, that since it's now ambiguous whether you mean Foo.apply or Foo.type, you have to just specify. Wanted to make sure I wasn't missing something. Thanks!
– user435779
Nov 15 '18 at 16:27
It's notFoo.type, it'sFoo. You are attempting to pass theobject Foo. You can't pass a type as a parameter (types are compile time only), those are two different things. Anyways, I am happy I could help.
– Markus Appel
Nov 15 '18 at 16:37
add a comment |
Thanks. After messing with it, that was also my conclusion, that since it's now ambiguous whether you mean Foo.apply or Foo.type, you have to just specify. Wanted to make sure I wasn't missing something. Thanks!
– user435779
Nov 15 '18 at 16:27
It's notFoo.type, it'sFoo. You are attempting to pass theobject Foo. You can't pass a type as a parameter (types are compile time only), those are two different things. Anyways, I am happy I could help.
– Markus Appel
Nov 15 '18 at 16:37
Thanks. After messing with it, that was also my conclusion, that since it's now ambiguous whether you mean Foo.apply or Foo.type, you have to just specify. Wanted to make sure I wasn't missing something. Thanks!
– user435779
Nov 15 '18 at 16:27
Thanks. After messing with it, that was also my conclusion, that since it's now ambiguous whether you mean Foo.apply or Foo.type, you have to just specify. Wanted to make sure I wasn't missing something. Thanks!
– user435779
Nov 15 '18 at 16:27
It's not
Foo.type, it's Foo. You are attempting to pass the object Foo. You can't pass a type as a parameter (types are compile time only), those are two different things. Anyways, I am happy I could help.– Markus Appel
Nov 15 '18 at 16:37
It's not
Foo.type, it's Foo. You are attempting to pass the object Foo. You can't pass a type as a parameter (types are compile time only), those are two different things. Anyways, I am happy I could help.– Markus Appel
Nov 15 '18 at 16:37
add a comment |
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