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On 09/07/2024 15:31, David Brown wrote:
> On 08/07/2024 19:39, BGB wrote:

>> Though, this one seems to be a common point of divergence between 
>> "SysV" and "Microsoft" ABIs. Sometimes a target will have an ABI 
>> defined, and the MS version was almost the same, just typically 
>> differing in that it passes structs by reference and provides a spill 
>> space for register arguments.
>>
> 
> I don't think it is helpful that you keep mixing /logical/ terms with 
> /implementation/ terms.
> 
> In C, there is no "pass by reference" or "return by reference".  It is 
> all done by value.

Arrays are passed by reference:

   void F(int a[20]) {}

   int main(void) {
     int x[20];
     F(x);
   }

Although the type of 'a' inside 'F' will be int* rather than int(*)[20].

> So if you have these structs and declarations :
> 
> struct small { uint16_t a; uint16_t b; };
> struct big { uint32_t xs[10]; };
> 
> struct small foos(struct small y);
> struct big foob(struct big y);
> 
> Then compilers will typically implement "x = foos(y)" as though it were:
> 
>      extern uint32_t foos(uint32_t ab);
>      uint32_t _1 = foos(y.a << 16) | (y.b);
>      struct small x = { _1 >> 16, _1 & 0xffff };
> 
> And they will typically implement "x = foosb(y)" as though it were:
> 
>      extern void foob(struct big * ret, const struct big * xs);
>      struct big x;
>      foob(&x, &y);

 From what I've seen, structs that are not small enough to be passed in 
registers, are copied to a temporary, and the address of that temporary 
is passed.

This seems to be the case even when the struct param is marked 'const'.

(My compiler won't create a copy when the parameter is 'const'. I 
assumed that was how gcc did it; I was wrong.)

This is for Win64 ABI, however an ABI will only say they are passed by 
reference; it will not stipulate making a copy. That is up to the 
language implementation.