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Re: [std-proposals] Copy-construct, move-construct, and PR-construct

From: Frederick Virchanza Gotham <cauldwell.thomas_at_[hidden]>
Date: Mon, 21 Aug 2023 22:59:36 +0100
I reply to a Breno, Jason and Jonathan in series below.

Breno GuimarĂ£es wrote:
> int main()
> {
> std::optional<std::mutex> om;
> reemplace(om, FuncThatReturnsMutex);
> // om = FuncThatReturnsMutex();
> }

That's very similar to what I did on Page 3 of my paper on NRVO:


Jason McKesson wrote:
> Any solution to a C++ problem that starts with "let's add a new
> reference type to the language" is not a solution worth having.
> There's really no point in considering it further.

I don't want to introduce a new kind of reference. I don't want to be
able to do this:

    int ^^a = b;

The use of the unary '^^' operator would only be seen in the parameter
list of a function that takes one and only one argument -- usually a
constructor or an overloaded assignment operator.

Jason McKesson continued writing:
> However, if this were to be entertained further, then some explanation
> of what's actually happening here is in order. You say that your new
> reference type means that the "argument must be a prvalue?". OK, but
> then you say "we need control over when the prvalue gets generated ".
> This makes no sense. The fact that a variable must be initialized by a
> prvalue doesn't change how variables work. "when the prvalue gets
> generated" happened *before the function call*. If you want it to
> happen differently from that, then you need to explain what those
> differences are and how that works. Are you saying that the parameter
> captures the entire expression, leaving it unevaluated until some time
> *within* the function (which is not what it means for a variable to
> "be a prvalue")? OK, so how does that work? Must the function be
> inlined? If not, how can a compiler pass *arbitrary code* to a
> concrete function that has no idea how any particular parameter was
> initialized?

So let's consider the following:

    extern mutex FuncThatReturnsMutex(int,int,int,int,int);

    int main(void)
        optional<mutex> om;
        om = FuncThatReturnsMutex(arg1,arg2,arg3,arg4,arg5);

My idea is still half-baked and subject to floatation but here's a
suggestion of how this could be implemented with the System V x86_64
calling convention:
(Step 1) The 5 arguments are evaluated and then pushed onto the stack
(none of them are put in registers)
(Step 2) The member function 'operator=' is invoked with return
address pushed onto the stack, with the RDI register set to the
address of 'om', and the RSI register set to the address of a tiny
thunk-like function. The RDX register is also set to the address of
another tiny thunk-like function (all will be explained later).
(Step 3) The body of 'operator=' does whatever it wants until it
encounters the '__emplace' keyword, at which point it invokes the
first tiny thunk-like function whose job it is to copy the 5 arguments
from their original location on the stack into the appropriate
registers (RSI, RDX, RCX, R8, R9) to be passed as arguments to
'FuncThatReturnsMutex'. If the function takes more parameters than can
go into registers, then the remaining arguments are copied from their
original location on the stack to the top of the stack (yes there will
be two copies of the data on the stack -- but this will rarely happen
as there are plenty of registers).
(Step 4) After the first tiny thunk-like function returns, the
function 'FuncThatReturnsMutex' is invoked with the RDI register set
to the address of the member variable 'buf' which resides inside the
'optional' object, and so the mutex gets emplaced in 'buf'.
(Step 5) The function 'FuncThatReturnsMutex' returns (or throws an
exception... but for now for simplicity let's just consider what
happens when it returns normally)
(Step 6) The remainder of the body of 'operator=' is executed.
(Step 7) Right before 'operator=' returns, the second tiny thunk-like
function is invoked and its job is to call the destructors of all the
arguments that were pushed onto the stack in Step 1, and it also
increments the stack pointer to its original value
(Step 8) The function 'operator=' returns.

So there's one possible implementation. There would be a tiny little
bit of gymnastics with the stack pointer when invoking the first tiny
thunk-like function (because it needs to jump back to the return
address even though it has pushed more stuff onto the stack), but it's

Jonathan Wakely wrote:
> Just stop replying to him. He clearly has no intention of doing anything except
> throwing silly ideas out again and again and again. Engaging with him clearly
> doesn't discourage him. Maybe ignoring him will.

Reverse psychology is at times uncanny, particularly in the realms of
international technical discussion, but thank you Jonathan you have
succeeded in motivating me to take this further.

Received on 2023-08-21 21:59:49