Counterexample: a guard stored in a boolean (boolean bad = a || b || c; if (bad) return;) is not split, so every variable of the guard is listed

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Avaliação

Dificuldade
4/5
Tempo estimado
3-5 dias
Facilidade para iniciantes
52/100
Tipo de issue
Bug
Clareza
Claramente especificada
Status de atividade
Ativa
Stack de tecnologia
java
Domínio
compilers

Direção de pesquisa

Start by reading RefinementError.factsOf and how boolean definitions and negated facts are handled; the issue identifies linkedTerms and isKnown as relevant. Run the supplied Java reproducer and compare its counterexample with the expected timeout == -1. Done when the stored-boolean guard produces the expected minimal counterexample without unrelated guard variables or the generated name.

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Descrição

bug error messages

Impact: when a guard is stored in a boolean before the if, the counterexample lists values for every variable of the guard and for a generated name, instead of only the value that breaks the call. Seen on the branch of #460 (pending merge), which fixes this for an inline guard; on main the counterexample for this input has the same noise and also leaves out timeout (#453).

Minimal reproducer

import liquidjava.specification.Refinement;

public class Probe {
    static void connect(@Refinement("_ >= 0") int timeout) {}

    static void probe(int port, int retries, int timeout) {
        boolean bad = port < 0 || retries < 0 || timeout > 60000;
        if (bad) return;
        connect(timeout);
    }
}

Expected: Counterexample: timeout == -1, as with the inline guard if (port < 0 || retries < 0 || timeout > 60000) return;, which on #460's branch (712759fe) gives:

Refinement Error: !(port < 0 || retries < 0 || timeout > 60000) is not a subtype of timeout >= 0
 --> Counterexample: timeout == -1

Actual (#460's branch @ 712759fe):

Refinement Error: #fresh⁹ == (port < 0 || retries < 0 || timeout > 60000) && !#fresh⁹ is not a subtype of timeout >= 0
 7 |         boolean bad = port < 0 || retries < 0 || timeout > 60000;
 8 |         if (bad) return;
 9 |         connect(timeout);
   |         ^^^^^^^^^^^^^^^^^
10 |     }
11 | }
 --> Counterexample: timeout == -1 && port == 0 && retries == 0 && #fresh⁹ == false

main @ 0053074e prints the same Found with Counterexample: port == 0 && retries == 0 && #fresh⁹ == false.

Cause. On #460's branch, RefinementError.factsOf splits !(a || b) into !a and !b, but here found is #fresh == (a || b || c) && !#fresh. The definition is one fact that shares terms with everything, so linkedTerms pulls in port, retries and #fresh. isKnown only recognises x == v facts, so !#fresh does not hide #fresh == false. Substituting a boolean defined as x == e into !x (giving !e) before splitting would give the same facts as the inline guard. Separately, found names the local bad as #fresh⁹.

🤖 Generated with Claude Code

Linguagem predominante
Java
Estrelas
71
Forks
37
Merge médio
1d 19h
PRs com merge (30d)
46

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