f82df0b285
This comes from https://reviews.llvm.org/D153003 By @rsmith, the test case is valid since: > Per [temp.type]/1.4 (http://eel.is/c++draft/temp.type#1.4), > >> Two template-ids are the same if [...] their corresponding template >> template-arguments refer to the same template. > so B<A> and B<NS::A> are the same type. The stricter "same sequence of > tokens" rule doesn't apply here, because using-declarations are not > definitions. > we should either (preferably) be including only the syntactic form of > the base specifier (because local syntax is what the ODR covers), or > the canonical type (which should be the same for both > using-declarations). Here we adopt the second suggested solutions. Reviewed By: cor3ntin, v.g.vassilev Differential Revision: https://reviews.llvm.org/D154324
45 lines
863 B
C++
45 lines
863 B
C++
// RUN: rm -rf %t
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// RUN: mkdir %t
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// RUN: split-file %s %t
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//
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// RUN: %clang_cc1 -std=c++20 -emit-module-interface -I%t %t/module1.cppm -o %t/module1.pcm
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// RUN: %clang_cc1 -std=c++20 -emit-module-interface -I%t %t/module2.cppm -o %t/module2.pcm
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// RUN: %clang_cc1 -std=c++20 -fprebuilt-module-path=%t %t/merge.cpp -verify -fsyntax-only
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//--- header.h
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namespace NS {
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template <int I>
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class A {
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};
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template <template <int I_> class T>
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class B {
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};
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}
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//--- module1.cppm
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// inside NS, using C = B<A>
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module;
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export module module1;
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#include "header.h"
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namespace NS {
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using C = B<A>;
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}
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export struct D : NS::C {};
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//--- module2.cppm
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// inside NS, using C = B<NS::A>
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module;
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export module module2;
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#include "header.h"
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namespace NS {
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using C = B<NS::A>;
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}
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export struct D : NS::C {};
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//--- merge.cpp
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// expected-no-diagnostics
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import module1;
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import module2;
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D d;
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