Skip to content

[C++20] [Modules] Try to treat B<A> and B<NS::A> as the same type in using declarations #63595

Closed
@ChuanqiXu9

Description

@ChuanqiXu9

This comes from https://reviews.llvm.org/D153003 and revealed by @zygoloid.

The reproducer:

// RUN: rm -rf %t
// RUN: mkdir %t
// RUN: split-file %s %t
//
// RUN: %clang_cc1 -std=c++20 -emit-module-interface -I%t %t/module1.cppm -o %t/module1.pcm
// RUN: %clang_cc1 -std=c++20 -emit-module-interface -I%t %t/module2.cppm -o %t/module2.pcm
// RUN: %clang_cc1 -std=c++20 -fprebuilt-module-path=%t %t/merge.cpp -verify -fsyntax-only

//--- header.h
namespace NS {
template <int I>
class A {
};

template <template <int I_> class T>
class B {
};
}

//--- module1.h
namespace NS {
using C = B<A>;
}
struct D : NS::C {};

//--- module1.cppm
// inside NS, using C = B<A>
module;
#include "header.h"
#include "module1.h"
export module module1;
export using ::D;

//--- module2.h
namespace NS {
using C = B<NS::A>;
}
struct D : NS::C {};

//--- module2.cppm
// inside NS, using C = B<NS::A>
module;
#include "header.h"
#include "module2.h"
export module module2;
export using ::D;

//--- merge.cpp
// expected-no-diagnostics
import module1;
import module2;
D d;

The explanation is: [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.

Metadata

Metadata

Assignees

Labels

clang:modulesC++20 modules and Clang Header Modules

Type

No type

Projects

No projects

Milestone

No milestone

Relationships

None yet

Development

No branches or pull requests

Issue actions