Do this program have the specified behavior?

[olcott]

#include <stdint.h>
typedef void (*ptr)();

int H(ptr x, ptr y)
{
x(y); // direct execution of P(P)
return 1;
}

int P(ptr x)
{
H(x, x);
return 1;
}

int main(void)
{
H(P, P);
}

For every H that simulates or executes its input and aborts or does not
abort its input P never reaches its last instruction.


--
Copyright 2021 Pete Olcott

Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer

[wij]

I would guess yes, it is specified: BAD, illegal program

[]$ g++ t.cpp
t.cpp: In function ‘int H(ptr, ptr)’:
t.cpp:6:2: error: too many arguments to function
6 | x(y); // direct execution of P(P)
| ~^~~
t.cpp: In function ‘int main()’:
t.cpp:18:3: error: invalid conversion from ‘int (*)(ptr)’ {aka ‘int (*)(void (*)())’} to ‘ptr’ {aka ‘void (*)()’} [-fpermissive]
18 | H(P, P);
| ^
| |
| int (*)(ptr) {aka int (*)(void (*)())}
t.cpp:4:11: note: initializing argument 1 of ‘int H(ptr, ptr)’
4 | int H(ptr x, ptr y)
| ~~~~^
t.cpp:18:6: error: invalid conversion from ‘int (*)(ptr)’ {aka ‘int (*)(void (*)())’} to ‘ptr’ {aka ‘void (*)()’} [-fpermissive]
18 | H(P, P);
| ^
| |
| int (*)(ptr) {aka int (*)(void (*)())}
t.cpp:4:18: note: initializing argument 2 of ‘int H(ptr, ptr)’
4 | int H(ptr x, ptr y)
| ~~~~^

[olcott]

On 11/16/2021 10:10 PM, wij wrote:
> On Wednesday, 17 November 2021 at 11:33:49 UTC+8, olcott wrote:

#include <stdint.h>

#include <stdio.h>
typedef int (*ptr)();

int H(ptr x, ptr y)
{
x(y); // direct execution of P(P)
return 1;
}

int P(ptr x)
{
H(x, x);
return 1;
}

int main(void)
{
H(P, P);
}

For every H that simulates or executes its input and aborts or does not
abort its input P never reaches its last instruction.

> I would guess yes, it is specified: BAD, illegal program

It must be compiled as C and the typedef was incorrect.
I can't find a way to specify the "C" calling convention for g++

[Chris M. Thomasson]

On 11/16/2021 8:46 PM, olcott wrote:
> #include <stdint.h>
> #include <stdio.h>
> typedef int (*ptr)();
>
> int H(ptr x, ptr y)
> {
>   x(y); // direct execution of P(P)
>   return 1;
> }
>
> int P(ptr x)
> {
>   H(x, x);
>   return 1;
> }
>
> int main(void)
> {
>   H(P, P);
> }

Stack overflow...

[Richard Damon]

On 11/16/21 11:46 PM, olcott wrote:
> On 11/16/2021 10:10 PM, wij wrote:
>> On Wednesday, 17 November 2021 at 11:33:49 UTC+8, olcott wrote:
>
> #include <stdint.h>
> #include <stdio.h>
> typedef int (*ptr)();
>
> int H(ptr x, ptr y)
> {
>   x(y); // direct execution of P(P)
>   return 1;
> }
>
> int P(ptr x)
> {
>   H(x, x);
>   return 1;
> }
>
> int main(void)
> {
>   H(P, P);
> }
>
> For every H that simulates or executes its input and aborts or does not
> abort its input P never reaches its last instruction.
>
>> I would guess yes, it is specified: BAD, illegal program
>
> It must be compiled as C and the typedef was incorrect.
> I can't find a way to specify the "C" calling convention for g++
>

Shows you don't understand C++.

Making it 'C' calling convention doesn't change the syntax for the call.

In C++, an empty parameter list specifies a void parameter list, while
in C it specifies tha mostly obsolete concept that this is a
non-prototype declaration that doesn't specify the parameters.

Ultimately the problem is that this runs into the issue that it is
impossible to express in C a function that takes or returns a pointer to
the same type as the function itself is, as that requires self
references in the type definition.

[olcott]

On 11/17/2021 2:28 AM, Mark Bluemel wrote:

> Surely a compiler can optimise this to remove the recursion, and for that matter, remove any functionality at all, as the code has no observable behaviour, as far as I can see (I'd be happy to be corrected on this!).
>

This is the question:

For every H that simulates or executes its input and aborts or does not

abort its input does any P ever reach its last instruction?


> $DIETY only knows what olcott was trying to show with this code.

[olcott]

This code does compile and run correctly.
The only difference from Ben's code is that
void (*ptr)(); was changed to int (*ptr)();

In comp.lang.c On 11/11/2021 2:31 PM, Ben Bacarisse:
transformed my syntax into his syntax

[Anurag Ranjan]

On 17/11/2021 04:46, Wolcott wrote:
> On 11/16/2021 10:10 PM, wij wrote:
>> On Wednesday, 17 November 2021 at 11:33:49 UTC+8, olcott wrote:
>
>
>

#include <stdio.h>

void count(int n)
{
static int d = 1;
printf("n = %d\n", n);
printf("d = %d\n", d);

d++;

if (n > 1)
{
count(n - 1);
}
printf("d = %d\n", d);
}
int main()
{
count(5);
return 0;
}

[Juha Nieminen]

In comp.lang.c++ olcott <No...@nowhere.com> wrote:
>>> typedef int (*ptr)();
>>>
>>> int H(ptr x, ptr y)
>>> {
>>> x(y); // direct execution of P(P)
>>> return 1;
>>> }
>

> This code does compile and run correctly.

It might compile as C, but it shouldn't compile as C++, because it's trying
to call an int(*)() with a parameter, even though that function (that the
pointer is pointing to does not take any parameter.)

Perhaps if it were an int(*)(void*) and then you reinterpret-cast the
parameter in order to call it, then it would compile.

As for "runs correctly", I don't see how this "runs correctly". Unless you
mean an infinite recursion that does nothing "runs correctly".

[Ben Bacarisse]

Juha Nieminen <nos...@thanks.invalid> writes:

> In comp.lang.c++ olcott <No...@nowhere.com> wrote:
>>>> typedef int (*ptr)();
>>>>
>>>> int H(ptr x, ptr y)
>>>> {
>>>> x(y); // direct execution of P(P)
>>>> return 1;
>>>> }
>>
>> This code does compile and run correctly.
>
> It might compile as C, but it shouldn't compile as C++, because it's trying
> to call an int(*)() with a parameter, even though that function (that the
> pointer is pointing to does not take any parameter.)

Just as it must not compile in C++, this code must compile in C. The
meaning of empty ()s in a function declaration is different between the
two languages.

When I tidied up the code, I wrote C, not C++, because C has a more
suitable type available.

--
Ben.

[olcott]

This is the question that I am asking:

Does this program have the specified behavior?

For every H that simulates or executes its input and aborts or does not
abort its input P never reaches its last instruction.

[olcott]

You did a really great job.