How do I convert msg.obj into the corresponding object and the compiler does not issue a warning?

Question

MutableList use as keyword

`when(msg?.what){
    MESSAGE_GET_LIST -> {
     if (msg.obj != null && msg.obj is MutableList<*>){
                        list = msg.obj as MutableList<BaseIncomeBean>
                        `

when I received a message and want to use it ,I must covert to MutableList use as keyword

Finally build my project and appear following warning

Warning:(51, 40) Unchecked cast: Any! to MutableList<BaseIncomeBean>

How do I fix this warning?

Please attach the code you're asking about as code, not as screenshots, and please try to ask your question in more understandable English. — yole, Apr 18 '18 at 07:18
If you are sure the list will always have this type you can ignore this warning. — user221256, Apr 18 '18 at 07:23

DummytoDummy · Answer 1 · 2018-04-18T07:34:32.503

Can you change if statement :

if(msg.arg1 == AsyncActionCallback.RESULT_OK && msg.obj != null && msg.obj is MutableList<BaseIncomeBean>)

from MutableList<*> to MutableList

"*" mean any in kotlin, Object in java. If you return an object reference, you must check whether the returned value is the class reference before casting another class. Because the return value is marked as an object, a different class reference may be returning. The IDE does not know that.

score 0 · Accepted Answer · answered Apr 18 '18 at 07:39

Kotlin has smart casts. But Generics are erased at runtime. Thus you never know the type of list contents. It would work to just cast it to MutableList<*>:

if (msg.obj is MutableList<*>)
    list = msg.obj as MutableList<*>

But since you set up your project, you know, that the cast should be successful. In this case you could just tell the compiler to ignore the warning:

if (msg.obj is MutableList<*>)
    @Suppress("UNCHECKED_CAST")
    list = msg.obj as MutableList<BaseIncomeBean>

How do I convert msg.obj into the corresponding object and the compiler does not issue a warning?

2 Answers2