Missing code implementation

[srikanth-sankaran]

From here: #1045 (comment)

import java.util.function.Supplier;

public class X {
	public static void main(String[] args) {
		String a = "Hello";
		if (((Supplier<String>) () -> a) instanceof Supplier)
		    System.out.print("yes");
	  	if (((Supplier<String>) () -> a) != null)
		    System.out.print("yes");

	}
}

results in Missing code implementation in the compiler error

[stephan-herrmann]

Seeing (((Supplier<String>) () -> a) != null) we want to raise "Redundant null check: this expression cannot be null" but in ProblemReporter.expressionNonNullComparison(Expression, boolean) nobody expects the expression to be a Lambda!! (after peeling off the cast, i.e.)
:)

[iloveeclipse]

@stephan-herrmann : the new test org.eclipse.jdt.core.tests.compiler.regression.NullReferenceTest.testGH1755() fails on all Java 17 builds with 1.7 compliance, like in https://download.eclipse.org/eclipse/downloads/drops4/I20240414-1800/testresults/html/org.eclipse.jdt.core.tests.compiler_ep432I-unit-cen64-gtk3-java17_linux.gtk.x86_64_17.html

junit.framework.ComparisonFailure: Invalid problem log .
----------- Expected ------------
----------\n
1. ERROR in X.java (at line 8)\n
if (((Supplier<String>) () -> a) != null)\n
^^^^^^^^^^^^^^^^^^^^^^^^^^^^\n
Redundant null check: this expression cannot be null\n
----------\n

------------ but was ------------
----------\n
1. ERROR in X.java (at line 6)\n
if (((Supplier<String>) () -> a) instanceof Supplier)\n
^^^^^^^\n
Lambda expressions are allowed only at source level 1.8 or above\n
----------\n
2. ERROR in X.java (at line 6)\n
if (((Supplier<String>) () -> a) instanceof Supplier)\n
^^^^^^^\n
The target type of this expression must be a functional interface\n
----------\n
3. ERROR in X.java (at line 8)\n
if (((Supplier<String>) () -> a) != null)\n
^^^^^^^\n
Lambda expressions are allowed only at source level 1.8 or above\n
----------\n
4. ERROR in X.java (at line 8)\n
if (((Supplier<String>) () -> a) != null)\n
^^^^^^^\n
The target type of this expression must be a functional interface\n
----------\n