为什么不能在unique_ptr中使用QFuture？

Question

下面是我的示例代码，我使用std::vector<std::unique_ptr<std::string>>作为未来的结果。

#include "mainwindow.h"
#include <QLineEdit>
#include <QtConcurrent/QtConcurrent>
#include <vector>

MainWindow::MainWindow(QWidget *parent) :
        QMainWindow(parent) {
    auto model = new QLineEdit(this);
    this->setCentralWidget(model);

    auto watcher = new QFutureWatcher<std::vector<std::unique_ptr<std::string>>>(/*this*/);
    auto future  = QtConcurrent::run([this]() -> std::vector<std::unique_ptr<std::string>> {
        std::vector<std::unique_ptr<std::string>> res;
        for (int k = 0; k < 100; ++k) {
            auto str = std::make_unique<std::string>("Hi");
            res.push_back(std::move(str));
        }
        return res;
    });

    connect(watcher, &QFutureWatcher<std::vector<std::unique_ptr<std::string>>>::finished, this, [=]() {
        for (const auto &item : future.result()){
            model->setText(model->text() + QString::fromStdString(*item));
        }
        delete watcher;
    });
    watcher->setFuture(future);
}

MainWindow::~MainWindow() {

}

但这段代码不能编译。这是原木，

/Users/ii/QT/qt-everywhere-src-6.2.0-beta4/include/QtCore/qfuture.h:328:12: note: in instantiation of member function 'std::vector<std::unique_ptr<std::string>>::vector' requested here
    return d.resultReference(0);
           ^
/Users/ii/CLionProjects/simpleQT/mainwindow.cpp:22:40: note: in instantiation of function template specialization 'QFuture<std::vector<std::unique_ptr<std::string>>>::result<std::vector<std::unique_ptr<std::string>>, void>' requested here
        for (const auto &item : future.result()){
                                       ^
/Applications/Xcode.app/Contents/Developer/Platforms/MacOSX.platform/Developer/SDKs/MacOSX12.1.sdk/usr/include/c++/v1/__memory/base.h:103:16: note: candidate template ignored: substitution failure [with _Tp = std::unique_ptr<std::string>, _Args = <std::unique_ptr<std::string> &>]: call to implicitly-deleted copy constructor of 'std::unique_ptr<std::string>'
constexpr _Tp* construct_at(_Tp* __location, _Args&& ...__args) {
               ^

Answer 1

为了只使用移动类型，您需要使用takeResult而不是result。

QFuture并不是真正考虑到可移动类型的设计，您最好使用可复制的类型或std::future。