2.3 control flow . cell 8 . the exercise of Polynomial Circuit about chisel-bootcamp HOT 7 CLOSED

It seems like the return type of your poly() is AnyVal, which suggests a possible bug in the poly() function.

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yes that makes sense, Ive found the problem: in the former scala numeric exercise, I used "if(select==0)else if(select==1)else if(select==2)" instead of "if(select==0)else if(select==1)else ..." to do the conditional branch. it works now. but it seems not a problem of the return type of poly() ? To be honest I suppose "if(select==0)else if(select==1)else if(select==2)" and "if(select==0)else if(select==1)else ..." have almost no difference in this case, isn't it? even it fixes the problem, I still don't know why...moreover, in the former use of the function expect, expect(IO, val) is often the case.

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If you define a function that returns in some branches, not others, Scala will infer the result type to be AnyVal. For example:

def funcTest(sel: Int) = {
    if (sel == 5) {
        10
    }
}

is equivalent to:

def funcTest(sel: Int): AnyVal = {
    if (sel == 5) {
        10
    }
}

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emm, I see. so that means:
a) the type AnyVal != Val
b) if the conditions listed are not complete the function will be defined as AnyVal defaultly, instead of a normal Val.
correct?

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a) Val is not a Scala type and lowercase val is a keyword and cannot be used as a variable name
b) The return type if none of the if conditions are triggered is Unit, and Scala will determine the result type to be the common supertype of the branch (Int) and Unit, which is AnyVal. Note that AnyVal is much less useful and should generally be avoided. If you need to sometimes return a type, Scala has the Option type.

In other words:

def funcTest(sel: Int): AnyVal = {
    if (sel == 5) {
        10
    }
}

is identical to:

def funcTest(sel: Int): AnyVal = {
    if (sel == 5) {
        10
    } else {
        Unit
    }
}

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I believe this time I get it~thats really helpful!
thanks for your patience!

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Glad to hear it!

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