Re: how cast works?

Thiago Adams <thiago.adams@gmail.com> writes:

Em 8/9/2024 2:20 PM, David Brown escreveu:

On 09/08/2024 12:57, Thiago Adams wrote:

Em 8/8/2024 6:41 PM, Bart escreveu:

On 08/08/2024 21:34, Thiago Adams wrote:

On 08/08/2024 16:42, Keith Thompson wrote:

Thiago Adams <thiago.adams@gmail.com> writes:

On 07/08/2024 17:00, Dan Purgert wrote:

On 2024-08-07, Thiago Adams wrote:

[...]

How about floating point?

Floating point is a huge mess, and has a few variations for
encoding;
though I think most C implementations use the one from the
IEEE on 1985
(uh, IEEE754, I think?)

>
I didn't specify properly , but my question was more about floating
point registers. I think in this case they have specialized
registers.

>
Who is "they"?
>
Some CPUs have floating-point registers, some don't.  C says nothing
about registers.
>
What exactly is your question?  Is it not already answered by reading
the "Conversions" section of the C standard?
>

>
>
This part is related with the previous question about the origins
of integer promotions.
>
We don't have "char" register or signed/unsigned register. But I
believe we may have double and float registers. So float does not
need to be converted to double.
>
There is no specif question here, just trying to understand the
rationally behind the conversions rules.

>
The rules have little to do with concrete machines with registers.
>
Your initial post showed come confusion about how conversions
work. They are not performed 'in-place', any more than writing `a
+ 1` changes the value of `a`.
>
Take:
>
     int a; double x;
>
     x = (double)a;
>
The cast is implicit here but I've written it out to make it
clear. My C compiler produces intermediate code like this before
converting it to native code:
>
     push x   r64                   # r64 means float64
     fix      r64 -> i32
     pop  a   i32
>
I could choose to interprete this code just as it is. Then, in
this execution model, there are no registers at all, only a stack
that can hold data of any type.
>
The 'fix' instruction pops the double value from the stack,
converts it to int (which involves changing both the bit-pattern,
and the bit-width), and pushes it back onto the stack.
>
Registers come into it when running it directly on a real
machine. But you seem more concerned with safety and correctness
than performance, so there's probably no real need to look at
actual generated native code.
>
That'll just be confusing (especially if you follow the advice to
generate only optimised code).
>
>
>
>

This part was always clear to me:
>
"They  are not performed 'in-place', any more than writing `a + 1`
changes the value of `a`."
>
Lets take double to int.
>
In this case the bits of double needs to be reinterpreted (copied
to) int.
>
So the answer "how it works" can be
>
always/generally machine has a instruction to do this
>
or.. this is defined by the IIE ... standard as ...
>
>

It would be helpful if you made more of an effort to write clearly
here.   (We know you can do so when you want to.)  It is very
difficult to follow what you are referring to here - what is "this
case" here?  A conversion from a double to an int certainly does not
re-interpret or copy bits - like other conversions, it copies the
/value/ to the best possible extent given the limitations of the
types.

>
Everything is a bit mixed up, but I'll try to explain the part about
registers that I have in mind.

Why you keep talking about registers?  The C standard does not talk
about registers, and the promotion rules are based on which *operations*
are available, not what kinds of registers a given CPU happens to have.

In C, when you have an expression like char + char, each char is
promoted to int. The computation then occurs as int + int.

Right.  And the most important thing to keep in mind is that it works
that way *because the C standard says so*.

If you're looking for a rationale for this, it's probably because some
early C compilers were for target systems that didn't provide operations
on narrow types.  (The PDP-11 does have instructions that operate on
8-bit bytes, but the arithmetic instructions operate only on 16-bit
words.)  But the C standard rules apply to all conforming C
implementations.  If a target CPU happens to provide a 1-byte add
instruction, a C compiler can't use it unless it can prove that the
result is consistent with C semantics.

Knowing why the C standard says what it says can be interesting, but
it's not necessarily directly useful.

On the other hand, when you have float + float, it remains as float + float.

That's implementation-defined.  Read the "Characteristics of floating
types <float.h>" section of the C standard (it's 5.2.5.3.3 in the N3220
draft) and look for FLT_EVAL_METHOD.  If FLT_EVAL_METHOD (defined in
<float.h>) is 0, float operations are done in type float.  If it's 1,
float operations are done in type double.  If it's 2, all floating-point
operations are done in type long double.

My guess for this design is that computations involving char are done
using registers that are the size of an int.

Who says the target CPU even has registers, or that computations can't
be done directly on values in memory?

But, float + float is not promoted to double, so I assume that the
computer has specific float registers or similar operation
instructions for float.

Again, some CPUs have floating-point registers and some do not.  Those
that don't might store floating-point values in the same registers used
for integer values, applying different instructions to operate on them.
Of those that do have floating-point registers, some have registers that
can hold a double value (typically 64 bits); they may or may not be able
to treat a half-register as a 32-bot float value.

The rules in the C standard are, for the most part, based on the
capabilities of CPUs that existed when the standard was written, not
necessarily on modern CPUs (though there have been tweaks in later
editions).

Regarding the part about signed/unsigned registers and operations, I
must admit that I'm not sure. I was planning to check on Compiler
Explorer, but I haven't done that yet.
>
I can frame the question like this: Does the computer make a
distinction when adding signed versus unsigned integers? Are there
specific assembly instructions for signed versus unsigned operations,
covering all possible combinations?

Maybe.

What is your goal here?  Are you trying to understand the history behind
the rules given in the C standard, or trying to understand the rules
themselves, or both?

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
void Void(void) { Void(); } /* The recursive call of the void */