We have to cover all branches with all Control-Flow expressions in Kotlin?

问题

I looked at the docs from Kotlin website, there are only two Control-Flow expressions: if and when.

For `if`:

the expression is required to have an else branch

For `when`:

The else branch is evaluated if none of the other branch conditions are satisfied. If when is used as an expression, the else branch is mandatory, unless the compiler can prove that all possible cases are covered with branch conditions.

Question

So it seems that there is no way to make a Control-Flow expression without covering all branches, Is it right? If not, Is there any way to make a Control-Flow expression to miss some branches; If so, why?

Following code will occur if must have both main and 'else' branches if used as an expression

override fun onReceive(context: Context?, intent: Intent?) {
    intent?.let {
        if (it.action == MySDK.BROADCAST_ACTION_LOGIN) {
            mListener.get()?.loggedOn(LoggedOnUserInfo.IT)
        }else if (it.action == MySDK.BROADCAST_ACTION_LOGOUT) {
            // Occur 'if must have both main and 'else' branches if used as an expression'
            mListener.get()?.loggedOut(LoggedOutUserInfo())
        }
    }
}

But following code pass compile.....

override fun onReceive(context: Context?, intent: Intent?) {
    intent?.let {
        if (it.action == MySDK.BROADCAST_ACTION_LOGIN) {
            mListener.get()?.loggedOn(LoggedOnUserInfo.IT)
            context!!.unregisterReceiver(this) // only add this line to test.
        }else if (it.action == MySDK.BROADCAST_ACTION_LOGOUT) {
            mListener.get()?.loggedOut(LoggedOutUserInfo())
        }
    }
}

回答1:

The trick here is not to use the if as an expression. My guess is that you placed the if at a let block, which returns its last statement, thus using the "result" of the if, thus treating it as an expression.

I suggest throwing away the let function (it is useless anyway here):

override fun onReceive(context: Context?, intent: Intent?) {
    if(intent != null) {
        if (intent.action == MySDK.BROADCAST_ACTION_LOGIN) {
            mListener.get()?.loggedOn(LoggedOnUserInfo.IT)
        } else if (intent.action == MySDK.BROADCAST_ACTION_LOGOUT) {
            mListener.get()?.loggedOut(LoggedOutUserInfo())
        }
    }
}

Your second version compiles because context!!.unregisterReceiver(this) has a different type than mListener.get()?.loggedOut(LoggedOutUserInfo()), which makes the types mismatch and prevents using the if as an expression.

P.S.

Kotlin has quite a few powerful control structures. I personally prefer this version:

override fun onReceive(context: Context?, intent: Intent?) {
    intent ?: return
    when(intent.action) {
        MySDK.BROADCAST_ACTION_LOGIN -> mListener.get()?.loggedOn(LoggedOnUserInfo.IT)
        MySDK.BROADCAST_ACTION_LOGOUT -> mListener.get()?.loggedOut(LoggedOutUserInfo())
    }
}

回答2:

So it seems that there is no way to make a Control-Flow expression without covering all branches, Is it right?

Yes

In the second case

mListener.get()?.loggedOn(LoggedOnUserInfo.IT)
context!!.unregisterReceiver(this)

Is not an expression anymore, the whole if block is a statement. However, you can also provide else with Unit in a first case if you really need an expression:

if (it.action == MySDK.BROADCAST_ACTION_LOGIN) {
    mListener.get()?.loggedOn(LoggedOnUserInfo.IT)
} else if (it.action == MySDK.BROADCAST_ACTION_LOGOUT) {
    // Occur 'if must have both main and 'else' branches if used as an expression'
    mListener.get()?.loggedOut(LoggedOutUserInfo())
} else Unit

But it would be better to avoid this code because it is less readable.

来源：https://stackoverflow.com/questions/40839544/we-have-to-cover-all-branches-with-all-control-flow-expressions-in-kotlin

标签

kotlin

control-flow