Issues using the in-line assembler

[solidstate1991]

On Thursday, 5 April 2018 at 04:48:02 UTC, Basile B. wrote:
> The "this" seems to be in R11, so you have to apply the asm 
> syntax for accessing the members using 
> <Type>.offsetof.<member>[R11], example:
>
> ```
> class Foo
> {
>     double a = 123456;
>     extern(D) double foo()
>     {
>         asm
>         {
>             naked;
>             movq    XMM0, Foo.a.offsetof[R11];
>             ret;
>         }
>     }
> }
>
> void main()
> {
>     import std.stdio;
>     (new Foo).foo(0,0).writeln;
> }
> ```
>
> However i cant find any specification saying that R11 is "this".
> With a free function just pass the instance as param and 
> replace R11 by the register where the instance is passed.

It seems that the compiler lets it through if I change it like 
this:

asm @nogc{
	naked;
	movd		XMM1, dword ptr sX[EBP];
	pslldq		XMM1, 4;
	movss		XMM1, dword ptr sY[EBP];
	movq		XMM0, xy;
	paddd		XMM0, XMM1;	// [x,y] + [sX,sY]
	movq		XMM3, qword ptr xy0[EBP];
	psubd		XMM0, XMM3;	// ([x,y] + [sX,sY] - [x_0,y_0])
	movq		XMM1, qword ptr ac[EBP];
	movq		XMM2, qword ptr bd[EBP];
	pmuludq		XMM1, XMM0;	// [A,0,C,0] * ([x,y] + [sX,sY] - [x_0,y_0])
	psrlq		XMM1, 16;	// ([A,0,C,0] * ([x,y] + [sX,sY] - 
[x_0,y_0]))>>16
	movups		XMM4, XMM0;
	psrldq		XMM4, 4;
	pslldq		XMM0, 4;
	por			XMM4, XMM0;
	pmuludq		XMM2, XMM4; // [0,B,0,D] * ([x,y] + [sX,sY] - [x_0,y_0])
	psrlq		XMM2, 16;	// ([0,B,0,D] * ([x,y] + [sX,sY] - 
[x_0,y_0]))>>16
	paddq		XMM1, XMM2; // ([A,B,C,D] * ([x,y] + [sX,sY] - 
[x_0,y_0]))>>16
	punpckldq	XMM3, XMM7;
	paddq		XMM1, XMM3;	// ([A,B,C,D] * ([x,y] + [sX,sY] - 
[x_0,y_0]))>>16 + [x_0,y_0]
	movups		XMM0, XMM1;	// Convert 64 bit vectors into 32 bit ones
	psrldq		XMM0, 4;
	por			XMM0, XMM1;	
	ret			;
}
I wonder if I can return an int[2] in XMM0. I can do some 
modifications to either move the result to the stack first, or 
add an import to core.simd (which needs to be refactored 
completely) and instead make the return type int4 on SIMD enabled 
CPUs.