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From: bart <bc@freeuk.com>
Newsgroups: comp.lang.c
Subject: =?UTF-8?Q?Re=3A_technology_discussion_=E2=86=92_does_the_world_need?=
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Date: Tue, 9 Jul 2024 21:28:49 +0100
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On 09/07/2024 18:22, Ben Bacarisse wrote:
> bart <bc@freeuk.com> writes:
> 
>> On 09/07/2024 16:58, Ben Bacarisse wrote:
>>> bart <bc@freeuk.com> writes:
>>>
>>>> Arrays are passed by reference:
>>>>
>>>>     void F(int a[20]) {}
>>>>
>>>>     int main(void) {
>>>>       int x[20];
>>>>       F(x);
>>>>     }
>>> This is the sort of thing that bad tutors say to students so that they
>>> never learn C properly.  All parameter passing in C is by value.  All of
>>> it.  You just have to know (a) what the syntax means and (b) what values
>>> get passed.
>>
>> The end result is that a parameter declared with value-array syntax is
>> passed using a reference rather than by value.
>>
>> And it does so because the language says, not because the ABI requires
>> it. A 2-byte array is also passed by reference.
> 
> An address value is passed by value.  C has only one parameter passing
> mechanism.  You can spin it as much as you like, but C's parameter
> passing is simple to understand, provided learner tune out voices like
> yours.

Little about C's type system is simple. You're doing your students a 
disservice if you try and hide all the quirks.

There's a discontinuity in the type system when it comes to parameter 
types (T is is non-array typedef; U is an array typedef):

   Non-Param      Same type as Param

   T a            T a
   T a[10]        T *a
   T a[10][20]    T (*a)[20]

   U a            U *a
   U a[10]        U (*a)[?]         ? is the length of U's array
   U A[10][20]    U (*a)[20][?]

Any top-level array type in a parameter (that is, its description in 
English starts with 'array of'), is converted to a pointer type (now 
starts with 'pointer to'). But the lower levels are not affected.

If you were to try the same exercise in one of my languages, if would be 
a smaller table:

   Param/Non-Param,
   T is array/non-array

   T a
   T a[10]
   T a[10][20]

> Sorry, I missed what you wrote.  I don't know why even brought up int
> (*)[20] but I thought you were saying that was the type of a.

int(*)[20] would be the type of 'a' inside my function, IF it was 
properly passed by value instead of some weird combination.

If I define this function in one of my languages, where '&' means 
pass-by-reference:

   proc f([20]int32 &A)=
   end

and tell it to transpile to C, it produces (minus the name decoration):

   static void f(i32 (*a)[20]) {
   }