The fixture is P -> M and P -> A -> M, where M's own flag --wide changes a struct it exports. Loaded with --wide --split, P gives M its args without --wide while A passes them on whole: Build(P) has to fail, name the module and both paths, and leave no bin/ anywhere. Loaded with --wide alone, both paths give M the same list, there is one M, and the build has to go through. Before the fix the first case built cleanly. Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
24 lines
792 B
C++
24 lines
792 B
C++
// SPDX-License-Identifier: LGPL-3.0-only
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// SPDX-FileCopyrightText: Copyright (C) 2026 Catcrafts®
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import std;
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import Crafter.Build;
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namespace fs = std::filesystem;
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using namespace Crafter;
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// Loads M with the default, so M gets whatever A was given.
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extern "C" Configuration CrafterBuildProject(std::span<const std::string_view> args) {
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Configuration* m = LocalProject({ .projectFile = "../m/project.cpp" });
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Configuration cfg;
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cfg.path = "./";
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cfg.name = "A";
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cfg.outputName = "A";
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cfg.type = ConfigurationType::LibraryStatic;
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ApplyStandardArgs(cfg, args);
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cfg.dependencies = { m };
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std::array<fs::path, 1> ifaces = { "interfaces/A" };
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std::array<fs::path, 0> impls = {};
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cfg.GetInterfacesAndImplementations(ifaces, impls);
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return cfg;
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}
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