In the Qt Docs, at the bottom of QFutureWatcher's Detailed
Description section there's this statement:
Any QFuture<T> can be watched by a QFutureWatcher<void> as
well. This is useful if only status or progress information
is needed; not the actual result data.
That statement fits what I'm trying to do - I only need the
status/progress information. So I attempt the lines:
QFutureWatcher<void> watcher;
watcher.setFuture(QtConcurrent::mappedReduced(list, mapped, reduce));
This fails to compile with the error message:
No viable conversion from 'QFuture<int>' to 'const QFuture<void>'
Where my mapped() function has the form
U function(const T &t);
And my reduce() function has the form
void function(int &result, const U &intermediate)
If I explicitly cast the mappedReduced() return value to a QFuture<void>:
watcher.setFuture(static_cast<QFuture<void>>(
QtConcurrent::mappedReduced(list, mapped, reduce)));
It compiles and seems to run fine, but I just wanted to double
check this was intended behavior, or if there was a better way
to do it? From the line in the docs, I assumed that the original
assignment was fine and I wouldn't need to explicitly cast.
Sean
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