Jorijn van der Graaf
3d2023cb5c
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>
12 lines
366 B
C++
12 lines
366 B
C++
export module M;
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import std;
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// --wide gives S more members, so the two builds of M the test asks for
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// differ in what they declare.
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#ifdef WIDE
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export struct S { std::int32_t a; std::int32_t b; std::int64_t pad[4]; };
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#else
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export struct S { std::int32_t a; std::int32_t b; };
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#endif
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export std::int32_t Size() {
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return static_cast<std::int32_t>(sizeof(S));
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}
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