Re: Order of evaluation of expression
Thank you, Doug.
"Doug Harrison [MVP]" wrote:
Call operator<< f and put it in functional notation:
f(f(f(wcout, Latch), "1"), "2");
Certainly, Latch will have been evaluated before the inner "f" is entered,
and the "output" of Latch will have completed before the output of "1"
begins, because the result of the former's "f" is required to call the
latter's "f". The necessary sequence points are all there (function
entry/exit), the Latch object lives on 'til the end of the full-expression
(the whole statement), and so this is OK.
I did not appreciate the three function calls were actually sequence points
but in that case it should certainly be all right then.
Assuming we're talking about a mutex, it's more usual to name your classes
Mutex and Lock instead of Lock and Latch, and your Mutex object would be
global or otherwise available to other threads. Instead of playing these
manipulator games, I would just write:
Lock lk(mx);
std::wcout << L"Something else 1.\n" << L"Something else 2.\n";
If you need finer control of the lifetime of lk, you can use:
// BLOCK
{
Lock lk(mx);
std::wcout << L"Something else 1.\n" << L"Something else 2.\n";
}
I think it's clearer this way. It's what you have to do with just about all
other classes.
It is actually a critical section (that is the way it was).
In my code, I do use your second version when output takes several lines
(separated by ; ). I have not used this manipulator yet but I was not going
to leave it the way I showed. My preferred way will be an equivalent of this:
inline std::wostream& wlog(std::wostream& os = std::wcout, const Latch& =
Latch(global_stream_lock))
{
return os;
}
and then
int main()
{
wlog() << L"Something 1" << ...
}
for short output operations. wlog() will potentially take other parameters
as well.