Java – How to pass class name in function arguments

java

I want to create a function that will perform some operation(most time occurring)
I created function like following

public void doSth()
{
   //logic
   ClassName.staticMethod();
   //logic
}

In My application there are many times this function will be called. Only the particular line will be change. I decided to give a common function.

Now my question is: How do I pass the ClassName in function arguments so that function body use it dynamically?

Thanks

Best Answer

You can do that, via Class.forName which accepts a fully-qualified class name and returns a Class instance. Then you have to get the Method for the static method in question via getMethod, and invoke it via invoke.

But passing around class names as strings is a suspect design decision. I'd look at alternatives, such as using singletons rather than static methods and an interface, that sort of thing.

Related Solutions

Java – How to test a class that has private methods, fields or inner classes

Update:

Some 10 years later perhaps the best way to test a private method, or any inaccessible member, is via @Jailbreak from the Manifold framework.
@Jailbreak Foo foo = new Foo();
// Direct, *type-safe* access to *all* foo's members
foo.privateMethod(x, y, z);
foo.privateField = value;
This way your code remains type-safe and readable. No design compromises, no overexposing methods and fields for the sake of tests.

If you have somewhat of a legacy Java application, and you're not allowed to change the visibility of your methods, the best way to test private methods is to use reflection.

Internally we're using helpers to get/set private and private static variables as well as invoke private and private static methods. The following patterns will let you do pretty much anything related to the private methods and fields. Of course, you can't change private static final variables through reflection.

Method method = TargetClass.getDeclaredMethod(methodName, argClasses);
method.setAccessible(true);
return method.invoke(targetObject, argObjects);

And for fields:

Field field = TargetClass.getDeclaredField(fieldName);
field.setAccessible(true);
field.set(object, value);

Notes:
1. TargetClass.getDeclaredMethod(methodName, argClasses) lets you look into private methods. The same thing applies for getDeclaredField.
2. The setAccessible(true) is required to play around with privates.

Java – Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.

Best Answer

Related Solutions

Java – How to test a class that has private methods, fields or inner classes

Java – Is Java “pass-by-reference” or “pass-by-value”

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