Type inference failed Kotlin rxjava map use

Multi tool use
Trying to add a compositeDisposable. using the code that follows:
compositeDisposable.add(
animalsObservable
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.filter s -> s.toLowerCase().startsWith("c")
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
.subscribeWith(animalsObserverAllCaps)
)
}
However, I get the following error:
Type inference failed: fun map(p0: Function!): Observable!
cannot be applied to
()
Type mismatch: inferred type is but Function! was expected
One type argument expected for interface Function

add a comment |
Trying to add a compositeDisposable. using the code that follows:
compositeDisposable.add(
animalsObservable
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.filter s -> s.toLowerCase().startsWith("c")
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
.subscribeWith(animalsObserverAllCaps)
)
}
However, I get the following error:
Type inference failed: fun map(p0: Function!): Observable!
cannot be applied to
()
Type mismatch: inferred type is but Function! was expected
One type argument expected for interface Function

add a comment |
Trying to add a compositeDisposable. using the code that follows:
compositeDisposable.add(
animalsObservable
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.filter s -> s.toLowerCase().startsWith("c")
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
.subscribeWith(animalsObserverAllCaps)
)
}
However, I get the following error:
Type inference failed: fun map(p0: Function!): Observable!
cannot be applied to
()
Type mismatch: inferred type is but Function! was expected
One type argument expected for interface Function

Trying to add a compositeDisposable. using the code that follows:
compositeDisposable.add(
animalsObservable
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.filter s -> s.toLowerCase().startsWith("c")
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
.subscribeWith(animalsObserverAllCaps)
)
}
However, I get the following error:
Type inference failed: fun map(p0: Function!): Observable!
cannot be applied to
()
Type mismatch: inferred type is but Function! was expected
One type argument expected for interface Function


edited Nov 15 '18 at 12:35
George
asked Nov 15 '18 at 12:26
GeorgeGeorge
3711318
3711318
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I would expect it to be as simple as:
.map it.toUpperCase()
Instead of
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
But anyway you're probably not using io.reactivex.functions.Function
, hence the error. If you really don't want to use a lambda then:
.map(object : io.reactivex.functions.Function<String, String>
override fun apply(t: String): String
return t.toUpperCase()
)
Or just import io.reactivex.functions.Function
and then use Function
.
unfortunately, not. it ends up having an error of s as is unresolved.
– George
Nov 15 '18 at 13:06
yes my bad, the default name of the parameter isit
. I updated the answer.
– lelloman
Nov 15 '18 at 13:08
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I would expect it to be as simple as:
.map it.toUpperCase()
Instead of
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
But anyway you're probably not using io.reactivex.functions.Function
, hence the error. If you really don't want to use a lambda then:
.map(object : io.reactivex.functions.Function<String, String>
override fun apply(t: String): String
return t.toUpperCase()
)
Or just import io.reactivex.functions.Function
and then use Function
.
unfortunately, not. it ends up having an error of s as is unresolved.
– George
Nov 15 '18 at 13:06
yes my bad, the default name of the parameter isit
. I updated the answer.
– lelloman
Nov 15 '18 at 13:08
add a comment |
I would expect it to be as simple as:
.map it.toUpperCase()
Instead of
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
But anyway you're probably not using io.reactivex.functions.Function
, hence the error. If you really don't want to use a lambda then:
.map(object : io.reactivex.functions.Function<String, String>
override fun apply(t: String): String
return t.toUpperCase()
)
Or just import io.reactivex.functions.Function
and then use Function
.
unfortunately, not. it ends up having an error of s as is unresolved.
– George
Nov 15 '18 at 13:06
yes my bad, the default name of the parameter isit
. I updated the answer.
– lelloman
Nov 15 '18 at 13:08
add a comment |
I would expect it to be as simple as:
.map it.toUpperCase()
Instead of
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
But anyway you're probably not using io.reactivex.functions.Function
, hence the error. If you really don't want to use a lambda then:
.map(object : io.reactivex.functions.Function<String, String>
override fun apply(t: String): String
return t.toUpperCase()
)
Or just import io.reactivex.functions.Function
and then use Function
.
I would expect it to be as simple as:
.map it.toUpperCase()
Instead of
.map(object : Function<String, String>()
@Throws(Exception::class)
fun apply(s: String): String
return s.toUpperCase()
)
But anyway you're probably not using io.reactivex.functions.Function
, hence the error. If you really don't want to use a lambda then:
.map(object : io.reactivex.functions.Function<String, String>
override fun apply(t: String): String
return t.toUpperCase()
)
Or just import io.reactivex.functions.Function
and then use Function
.
edited Nov 15 '18 at 13:06
answered Nov 15 '18 at 12:50


lellomanlelloman
8,73133454
8,73133454
unfortunately, not. it ends up having an error of s as is unresolved.
– George
Nov 15 '18 at 13:06
yes my bad, the default name of the parameter isit
. I updated the answer.
– lelloman
Nov 15 '18 at 13:08
add a comment |
unfortunately, not. it ends up having an error of s as is unresolved.
– George
Nov 15 '18 at 13:06
yes my bad, the default name of the parameter isit
. I updated the answer.
– lelloman
Nov 15 '18 at 13:08
unfortunately, not. it ends up having an error of s as is unresolved.
– George
Nov 15 '18 at 13:06
unfortunately, not. it ends up having an error of s as is unresolved.
– George
Nov 15 '18 at 13:06
yes my bad, the default name of the parameter is
it
. I updated the answer.– lelloman
Nov 15 '18 at 13:08
yes my bad, the default name of the parameter is
it
. I updated the answer.– lelloman
Nov 15 '18 at 13:08
add a comment |
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