std:异步在某些msvc版本中失败，解决方法是什么？
std:异步在某些msvc版本中失败，解决方法是什么？
提问于 2022-11-02 16:16:26
以下代码：
#include <vector>
#include <algorithm>
#include <thread>
#include <future>
#include <iostream>
int main() {
std::vector<int> v(1000);
for(unsigned i = 0; i < v.size(); i++) {
v[i] = i;
}
int bb = 2;
std::cout << v.back() << std::endl;
auto f = [&](int& x) {x = 2*x; bb=4; };
std::async(std::launch::async, std::for_each<decltype(v.begin()),decltype(f)>,
v.begin(), v.end(), f);
std::cout << v.back() << std::endl;
std::cout << "bb: " << bb;
return 0;
}未能使用msvc版本< 19.32进行编译。例如，msvc = 19.29 (在godbolt.org上编译)的错误如下所示：
example.cpp
<source>(16): warning C4834: discarding return value of function with 'nodiscard' attribute
C:/data/msvc/14.31.31108/include\future(314): error C2280: 'main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37> &main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>::operator =(const main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37> &)': attempting to reference a deleted function
<source>(14): note: see declaration of 'main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>::operator ='
<source>(14): note: 'main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37> &main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>::operator =(const main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37> &)': function was explicitly deleted
C:/data/msvc/14.31.31108/include\future(309): note: while compiling class template member function 'void std::_Associated_state<_Ty>::_Set_value_raw(_Ty &&,std::unique_lock<std::mutex> *,bool)'
with
[
_Ty=main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>
]
C:/data/msvc/14.31.31108/include\future(305): note: see reference to function template instantiation 'void std::_Associated_state<_Ty>::_Set_value_raw(_Ty &&,std::unique_lock<std::mutex> *,bool)' being compiled
with
[
_Ty=main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>
]
C:/data/msvc/14.31.31108/include\future(722): note: see reference to class template instantiation 'std::_Associated_state<_Ty>' being compiled
with
[
_Ty=main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>
]
C:/data/msvc/14.31.31108/include\future(720): note: while compiling class template member function 'std::_State_manager<_Ty>::~_State_manager(void) noexcept'
with
[
_Ty=main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>
]
C:/data/msvc/14.31.31108/include\future(879): note: see reference to function template instantiation 'std::_State_manager<_Ty>::~_State_manager(void) noexcept' being compiled
with
[
_Ty=main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>
]
C:/data/msvc/14.31.31108/include\future(860): note: see reference to class template instantiation 'std::_State_manager<_Ty>' being compiled
with
[
_Ty=main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>
]
<source>(17): note: see reference to class template instantiation 'std::future<main::<lambda_fa7e30e7ff267c277ebbd82c8a7f9e37>>' being compiledhttps://godbolt.org/z/WsovKcnnc
它是用gcc、icc、clang和msvc (>=19.32) (使用-lpthread)成功编译的。
是否可以使用msvc版本= 19.29 (VS19)进行编译？
我尝试了/std:c++latest选项，但这并没有什么区别。
回答 1
Stack Overflow用户
发布于 2022-11-03 12:02:49
如果Marek认为MSVC的STL实现是正确的，那么您可以通过使用std::packaged_task和std：：线程来避免调用std：：异步。
#include <thread>
#include <future>
#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
int main()
{
std::vector<int> v(100);
std::iota(v.begin(), v.end(), 0);
int bb = -2;
auto f = [&] (int& x) {
x = 2 * x;
bb = 4;
};
using Iter = decltype(v.begin());
using Fun = decltype(f);
auto foreach = [] (Iter begin, Iter end, Fun f) {
std::for_each<Iter, Fun>(begin, end, f);
};
std::packaged_task<void(Iter, Iter, Fun)> task(foreach);
auto future = task.get_future();
std::thread t(std::move(task), v.begin(), v.end(), f);
future.get();
t.join();
//
std::cout << v.back() std::endl;
std::cout << "bb: " << bb << std::endl;
return 0;
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/74292454
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