Subtle call to inherited method¶

ID: java/subtle-inherited-call
Kind: problem
Security severity: 
Severity: warning
Precision: very-high
Tags:
   - quality
   - maintainability
   - readability
Query suites:
   - java-code-quality.qls
   - java-security-and-quality.qls

Click to see the query in the CodeQL repository

If a call is made to a method from an inner class A, and a method of that name is defined in both a superclass of A and an outer class of A, it is not clear to a programmer which method is intended to be called.

Example¶

In the following example, it is not clear whether the call to printMessage calls the method that is defined in Outer or Super.

public class Outer
{
	void printMessage() {
		System.out.println("Outer");
	}
	
	class Inner extends Super
	{
		void ambiguous() {
			printMessage();  // Ambiguous call
		}
	}
	
	public static void main(String[] args) {
		new Outer().new Inner().ambiguous();
	}
}

class Super
{
	void printMessage() {
		System.out.println("Super");
	}
}

Inherited methods take precedence over methods in outer classes, so the method in the superclass is called. However, such situations are a potential cause of confusion and defects.

Recommendation¶

Resolve the ambiguity by explicitly qualifying the method call:

To specify the outer class, prefix the method with Outer.this..
To specify the superclass, prefix the method with super.. In the above example, the call to printMessage could be replaced by either Outer.this.printMessage or super.printMessage, depending on which method you intend to call. To preserve the behavior in the example, use super.printMessage.

References¶

Inner Classes Specification: What are top-level classes and inner classes?.

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