Question

I have 2 alternatives to implement a calculation method, and I am wondering what would be the better approach.

The method needs some int and double parameters and (in some cases) a special flag to do some different calculation.

In the first example, I could call the method with ‘calculateFoo(1, 2.0d)’ to have the boolean flag == FALSE.

In the second example, I always have to set the boolean flag (even if I do not need it)

Method 1: (here I’m using the ‘…’ as an ‘method overloading’ parameter)

public SomeObject calculateFoo(int pIntValue, double pDoubleValue, boolean... pDoSpecialStuff) {

    if (pDoSpecialStuff.length == 0 || pDoSpecialStuff[0] == false) {
        // method was called without boolean flag or boolean flag was set to FALSE

        // do some calculation stuff
        // ...
    } else {
        // method was called with boolean flag == TRUE

        // do some other calculation stuff
        // ...
    }

    return SomeObject; 
}

Method 2: (this is the ‘common’ approach)

public SomeObject calculateFoo(int pIntValue, double pDoubleValue, boolean pDoSpecialStuff) {

    if (pDoSpecialStuff == false) {
        // method was called with boolean flag == FALSE

        // do some calculation stuff
        // ...
    } else {
        // method was called with boolean flag == TRUE

        // do some other calculation stuff
        // ...
    }

    return SomeObject; 
}

Answer 1

Consider the following pattern:

public ResultType calcFoo( int i, double d ) {
    return calc( i, d, false );
}

public ResultType calcFoo( int i, double d, boolean flag ) {
    if( flag ) {
        ...
        return result;
    }
    else {
        ...
        return result;
    }
}

Generally it’s better to use an enum instead of a boolean flag. It makes your code more readable and is just as fast.

I noticed you considered using varargs. If you want to use more flags consider using an EnumSet to pass a set of flags to the method. If you want to pass 0 or 1 flags, varargs is even more definitely an antipattern.