57b08b0944
to reflect the new license. These used slightly different spellings that defeated my regular expressions. We understand that people may be surprised that we're moving the header entirely to discuss the new license. We checked this carefully with the Foundation's lawyer and we believe this is the correct approach. Essentially, all code in the project is now made available by the LLVM project under our new license, so you will see that the license headers include that license only. Some of our contributors have contributed code under our old license, and accordingly, we have retained a copy of our old license notice in the top-level files in each project and repository. llvm-svn: 351648
104 lines
2.7 KiB
C++
104 lines
2.7 KiB
C++
//===----------------------------------------------------------------------===//
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//
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// Part of the LLVM Project, under the Apache License v2.0 with LLVM Exceptions.
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// See https://llvm.org/LICENSE.txt for license information.
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// SPDX-License-Identifier: Apache-2.0 WITH LLVM-exception
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//
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//===----------------------------------------------------------------------===//
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// <tuple>
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// template <class... Types> class tuple;
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// tuple& operator=(const tuple& u);
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// UNSUPPORTED: c++98, c++03
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#include <tuple>
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#include <memory>
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#include <string>
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#include <cassert>
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#include "test_macros.h"
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struct NonAssignable {
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NonAssignable& operator=(NonAssignable const&) = delete;
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NonAssignable& operator=(NonAssignable&&) = delete;
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};
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struct CopyAssignable {
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CopyAssignable& operator=(CopyAssignable const&) = default;
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CopyAssignable& operator=(CopyAssignable &&) = delete;
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};
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static_assert(std::is_copy_assignable<CopyAssignable>::value, "");
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struct MoveAssignable {
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MoveAssignable& operator=(MoveAssignable const&) = delete;
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MoveAssignable& operator=(MoveAssignable&&) = default;
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};
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int main()
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{
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{
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typedef std::tuple<> T;
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T t0;
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T t;
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t = t0;
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}
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{
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typedef std::tuple<int> T;
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T t0(2);
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T t;
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t = t0;
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assert(std::get<0>(t) == 2);
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}
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{
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typedef std::tuple<int, char> T;
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T t0(2, 'a');
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T t;
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t = t0;
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assert(std::get<0>(t) == 2);
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assert(std::get<1>(t) == 'a');
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}
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{
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typedef std::tuple<int, char, std::string> T;
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const T t0(2, 'a', "some text");
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T t;
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t = t0;
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assert(std::get<0>(t) == 2);
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assert(std::get<1>(t) == 'a');
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assert(std::get<2>(t) == "some text");
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}
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{
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// test reference assignment.
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using T = std::tuple<int&, int&&>;
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int x = 42;
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int y = 100;
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int x2 = -1;
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int y2 = 500;
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T t(x, std::move(y));
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T t2(x2, std::move(y2));
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t = t2;
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assert(std::get<0>(t) == x2);
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assert(&std::get<0>(t) == &x);
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assert(std::get<1>(t) == y2);
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assert(&std::get<1>(t) == &y);
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}
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{
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// test that the implicitly generated copy assignment operator
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// is properly deleted
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using T = std::tuple<std::unique_ptr<int>>;
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static_assert(!std::is_copy_assignable<T>::value, "");
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}
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{
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using T = std::tuple<int, NonAssignable>;
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static_assert(!std::is_copy_assignable<T>::value, "");
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}
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{
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using T = std::tuple<int, CopyAssignable>;
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static_assert(std::is_copy_assignable<T>::value, "");
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}
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{
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using T = std::tuple<int, MoveAssignable>;
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static_assert(!std::is_copy_assignable<T>::value, "");
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}
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}
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