[GROOVY-11005] CompileStatic: Cannot set default-visible property in superclass that has a getter without a setter

• Type: Bug
• Status: Closed
• Priority: Minor
• Resolution: Fixed
• Affects Version/s: 3.0.17
• Fix Version/s: 3.0.18
• Component/s: Compiler
• Assignee: Eric Milles
• Reporter: Jason Garrett
• Created: 07/Apr/23 21:02
• Updated: 23/Aug/23 09:09
• Resolved: 11/Apr/23 00:06

Description

If a class's ancestors declare a default-visibility property and a getter for that property:

class HasProperty {
    Object foo

    Object getFoo() {
       return foo
    }
} 

Then the class, if it is CompileStatic, cannot set that property.

 

This fails to compile:

import groovy.transform.CompileStatic

@CompileStatic
class SetsProperty extends HasProperty {
    void doStuff() {
       foo = "example"
    }
} 

Compiler output:

SetsProperty.groovy: 8: [Static type checking] - Cannot set read-only property: foo
 @ line 8, column 3.
           foo = "example"

 

It also fails to compile with "this.@":

import groovy.transform.CompileStatic

@CompileStatic
class SetsProperty extends HasProperty {
    void doStuff() {
       this.@foo = "example"
    }
} 

Results in:

SetsProperty.groovy: 8: [Static type checking] - The field HasProperty.foo is not accessible
 @ line 8, column 9.
           this.@foo = "example"

 

These do compile if the class is not CompileStatic, or if the property is explicitly declared "public".

 

 

Issue Links

• is related to: GROOVY-9127 Cannot set protected field from subclass
• is related to: GROOVY-5450 Final field assignment check on static compilation