Does Java support default parameter values?

Question

I came across some Java code that had the following structure:

public MyParameterizedFunction(String param1, int param2)
{
    this(param1, param2, false);
}         


        

Answer 1

No, but you can very easily emulate them. What in C++ was:

public: void myFunction(int a, int b=5, string c="test") { ... }

In Java, it will be an overloaded function:

public void myFunction(int a, int b, string c) { ... }

public void myFunction(int a, int b) {
    myFunction(a, b, "test");
}

public void myFunction(int a) {
    myFunction(a, 5);
}

Earlier was mentioned, that default parameters caused ambiguous cases in function overloading. That is simply not true, we can see in the case of the C++: yes, maybe it can create ambiguous cases, but these problem can be easily handled. It simply wasn't developed in Java, probably because the creators wanted a much simpler language as C++ was - if they had right, is another question. But most of us don't think he uses Java because of its simplicity.

Answer 2

No, but the simplest way to implement this is:

public myParameterizedFunction(String param1, int param2, Boolean param3) {

    param3 = param3 == null ? false : param3;
}

public myParameterizedFunction(String param1, int param2) {

    this(param1, param2, false);
}

or instead of the ternary operator, you can use if:

public myParameterizedFunction(String param1, int param2, Boolean param3) {

    if (param3 == null) {
        param3 = false;
    }
}

public myParameterizedFunction(String param1, int param2) {

    this(param1, param2, false);
}

Answer 3

Unfortunately, yes.

void MyParameterizedFunction(String param1, int param2, bool param3=false) {}

could be written in Java 1.5 as:

void MyParameterizedFunction(String param1, int param2, Boolean... params) {
    assert params.length <= 1;
    bool param3 = params.length > 0 ? params[0].booleanValue() : false;
}

But whether or not you should depend on how you feel about the compiler generating a

new Boolean[]{}

for each call.

For multiple defaultable parameters:

void MyParameterizedFunction(String param1, int param2, bool param3=false, int param4=42) {}

could be written in Java 1.5 as:

void MyParameterizedFunction(String param1, int param2, Object... p) {
    int l = p.length;
    assert l <= 2;
    assert l < 1 || Boolean.class.isInstance(p[0]);
    assert l < 2 || Integer.class.isInstance(p[1]);
    bool param3 = l > 0 && p[0] != null ? ((Boolean)p[0]).booleanValue() : false;
    int param4 = l > 1 && p[1] != null ? ((Integer)p[1]).intValue() : 42;
}

This matches C++ syntax, which only allows defaulted parameters at the end of the parameter list.

Beyond syntax, there is a difference where this has run time type checking for passed defaultable parameters and C++ type checks them during compile.

Answer 4

Instead of using:

void parameterizedMethod(String param1, int param2) {
    this(param1, param2, false);
}

void parameterizedMethod(String param1, int param2, boolean param3) {
    //use all three parameters here
}

You could utilize java's Optional functionality by having a single method:

void parameterizedMethod(String param1, int param2, @Nullable Boolean param3) {
    param3 = Optional.ofNullable(param3).orElse(false);
    //use all three parameters here
}

The main difference is that you have to use wrapper classes instead of primitive Java types to allow null input.Boolean instead of boolean, Integer instead of int and so on.

Answer 5

No. In general Java doesn't have much (any) syntactic sugar, since they tried to make a simple language.

Answer 6

As Scala was mentioned, Kotlin is also worth mentioning. In Kotlin function parameters can have default values as well and they can even refer to other parameters:

fun read(b: Array<Byte>, off: Int = 0, len: Int = b.size) {
    ...
}

Like Scala, Kotlin runs on the JVM and can be easily integrated into existing Java projects.