Interview question

Can you overload methods that differ only by generic type parameters? क्या आप methods को सिर्फ generic type parameters के अंतर से overload कर सकते हैं?

Answer

No - because of type erasure, two methods that differ only in their generic type parameter end up with the IDENTICAL erased signature at compile time, causing a 'duplicate method' compile error.

class Processor {
    public void process(List<String> list) {
        System.out.println('Processing strings');
    }

    // COMPILE ERROR: 'process(List<Integer>)' has the same erasure
    // as 'process(List<String>)' - after erasure, BOTH become
    // process(List) with no distinguishing information
    // public void process(List<Integer> list) {
    //     System.out.println('Processing integers');
    // }
}

// WORKAROUND 1: use different method names
class ProcessorFixed {
    public void processStrings(List<String> list) {
        System.out.println('Processing strings');
    }
    public void processIntegers(List<Integer> list) {
        System.out.println('Processing integers');
    }
}

// WORKAROUND 2: pass a Class<T> token to distinguish at runtime
class ProcessorWithToken {
    public <T> void process(List<T> list, Class<T> type) {
        System.out.println('Processing ' + type.getSimpleName());
    }
}
new ProcessorWithToken().process(List.of('a', 'b'), String.class);

// Note: methods DIFFERING in the RAW type (not just generic parameter)
// CAN coexist - this is true overloading, not affected by erasure
class ValidOverload {
    public void process(List<String> list) { }
    public void process(Set<String> set) { }  // OK - List vs Set are different raw types
}

नहीं - type erasure की वजह से, generic type parameter में अलग होने वाले दो methods compile time पर IDENTICAL erased signature पाते हैं, 'duplicate method' compile error आती है।

class Processor {
    public void process(List<String> list) {
        System.out.println('Strings process हो रहे हैं');
    }

    // COMPILE ERROR: erasure के बाद दोनों process(List) बन जाते हैं
    // public void process(List<Integer> list) {
    //     System.out.println('Integers process हो रहे हैं');
    // }
}

// WORKAROUND 1: अलग method names
class ProcessorFixed {
    public void processStrings(List<String> list) {
        System.out.println('Strings process हो रहे हैं');
    }
    public void processIntegers(List<Integer> list) {
        System.out.println('Integers process हो रहे हैं');
    }
}

// WORKAROUND 2: Class<T> token pass करें
class ProcessorWithToken {
    public <T> void process(List<T> list, Class<T> type) {
        System.out.println('Processing ' + type.getSimpleName());
    }
}
new ProcessorWithToken().process(List.of('a', 'b'), String.class);

// Raw type में अलग methods coexist कर सकते हैं
class ValidOverload {
    public void process(List<String> list) { }
    public void process(Set<String> set) { }  // ठीक है - List vs Set अलग raw types
}

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