register

[orium69]

hi everyone, i'm wondering if there is a way to have sure that a
variable is allocated in the cache, after its declaration with
"register"? Tks!

 

[Gordon Burditt]

hi everyone, i'm wondering if there is a way to have sure that a

variable is allocated in the cache, after its declaration with
"register"? Tks!

On all processors I have ever heard of with a cache, actual CPU
registers are never part of the cache. Actual CPU registers
are normally much faster than the cache.

You can never be sure that the keyword 'register' will affect
the code at all. And if you use it enough, you *WILL* run out
of actual CPU registers. If you could force a variable into a
CPU register, it might slow down the code because the register
might be better used as an invisible temporary. The compiler
is often smarter than you are.

Gordon L. Burditt

 

[Jordan Abel]

On all processors I have ever heard of with a cache, actual CPU
registers are never part of the cache. Actual CPU registers
are normally much faster than the cache.

You can never be sure that the keyword 'register' will affect
the code at all. And if you use it enough, you *WILL* run out
of actual CPU registers. If you could force a variable into a
CPU register, it might slow down the code because the register
might be better used as an invisible temporary. The compiler
is often smarter than you are.

The compiler also isn't required to listen to you. All "register" means
is that the compiler is free to not give the variable an address. This
could otherwise be determined by code analysis, i'm sure, but a keyword
is simpler.

 

[Kenny McCormack]

hi everyone, i'm wondering if there is a way to have sure that a
variable is allocated in the cache, after its declaration with
"register"? Tks!

The answers you will get will all boil down to "You can't know and you
shouldn't want to know."

However, the bottom line is: Invoke the compiler with an option that causes
it to output assembly (e.g., "gcc -S") and then look at the generated
assembly. You should be able to figure out what it did with your
"register" directive.

 

[Mark McIntyre]

On Wed, 09 Nov 2005 20:10:12 GMT, in comp.lang.c ,

The answers you will get will all boil down to "You can't know and you
shouldn't want to know."

However, the bottom line is: Invoke the compiler with an option that causes
it to output assembly (e.g., "gcc -S") and then look at the generated
assembly. You should be able to figure out what it did with your
"register" directive.

This won't tell you a thing about how the CPU will use the cache at
runtime. Never mind that if its in a register, it cannot be in a
cache...

In fact, this is a classic example of why answering offtopic questions
here is a bad idea.

 

[Alex Fraser]

[snip]

The compiler also isn't required to listen to you. All "register" means
is that the compiler is free to not give the variable an address.

I would say all it means is that you cannot take the address of the object.
What you wrote follows from that fact.

This could otherwise be determined by code analysis, i'm sure, but a
keyword is simpler.

These days, such analysis (and then some) is certainly practical, but
presumably it wasn't always that way, and that is why the keyword exists.

Alex

 

[Keith Thompson]

The compiler also isn't required to listen to you. All "register" means
is that the compiler is free to not give the variable an address. This
could otherwise be determined by code analysis, i'm sure, but a keyword
is simpler.

Here's what the standard says, C99 6.7.1p4:

A declaration of an identifier for an object with storage-class
specifier register suggests that access to the object be as fast
as possible. The extent to which such suggestions are effective is
implementation-defined.

with a footnote:

The implementation may treat any register declaration simply as an
auto declaration. However, whether or not addressable storage is
actually used, the address of any part of an object declared with
storage-class specifier register cannot be computed, either
explicitly (by use of the unary & operator as discussed in
6.5.3.2) or implicitly (by converting an array name to a pointer
as discussed in 6.3.2.1). Thus, the only operator that can be
applied to an array declared with storage-class specifier register
is sizeof.

6.5.4.2 says you can't apply unary "&" to a register-qualified object.
6.3.2.1 says that the implicit conversion of an array expression to a
pointer to its first element invokes undefined behavior if the
designated array object has register storage class. (I'm not sure why
it's undefined behavior rather than a constraint violation.)n

 

[Kenny McCormack]

This won't tell you a thing about how the CPU will use the cache at
runtime. Never mind that if its in a register, it cannot be in a
cache...

You're being needlessly pedantic - and, might I add, definitely not PC.
I think we can assume that the OP meant "CPU register" when he said
"cache". Remember, these dorks use terminology very loosely and generally
just repeat words they've overheard.

In fact, this is a classic example of why answering offtopic questions
here is a bad idea.

Sez you.

 

[Keith Thompson]

You're being needlessly pedantic - and, might I add, definitely not PC.
I think we can assume that the OP meant "CPU register" when he said
"cache".

Perhaps, but I don't think we *need* to assume any such thing. It's
likely that the OP didn't understand what registers and cache really
are. If he's been following this thread, he should now have a much
better understanding of the terms.

(The term "cache" as a generic term is applicable to registers, but in
this context it more commonly refers to something else.)

I suppose we could have guessed what he meant by the term "cache" and
pretended his usage was correct. Then we could have fooled him into
thinking that he was getting a real answer to his question, something
like "Just use the register keyword, it guarantees that the variable
will be stored in cache."

But most of us are more interested in passing on knowledge than in
playing stupid games like that.

 

[pete]

(e-mail address removed) (Kenny McCormack) writes:

But most of us are more interested in passing on knowledge than in
playing stupid games like that.

But Kenny isn't.

http://www.codecomments.com/Unix_Programming/message665290.html

<OT>
Poor Kenny. The people in comp.lang.c aren't amenable to redefining
the purpose of the newsgroup to suit him. One wonders why he continues
to hang around there ;-)
</OT>
I stick around, read, and post, for the same reasons I do in
any other joke newsgroup - because it is so much fun.

 

[Richard Bos]

You can never be sure that the keyword 'register' will affect
the code at all. And if you use it enough, you *WILL* run out
of actual CPU registers.

Erm... I spot an inconsistency.

Richard

 

[Simon Biber]

Here's what the standard says, C99 6.7.1p4:

A declaration of an identifier for an object with storage-class
specifier register suggests that access to the object be as fast
as possible. The extent to which such suggestions are effective is
implementation-defined.

with a footnote:

The implementation may treat any register declaration simply as an
auto declaration. However, whether or not addressable storage is
actually used, the address of any part of an object declared with
storage-class specifier register cannot be computed, either
explicitly (by use of the unary & operator as discussed in
6.5.3.2) or implicitly (by converting an array name to a pointer
as discussed in 6.3.2.1). Thus, the only operator that can be
applied to an array declared with storage-class specifier register
is sizeof.

6.5.4.2 says you can't apply unary "&" to a register-qualified object.
6.3.2.1 says that the implicit conversion of an array expression to a
pointer to its first element invokes undefined behavior if the
designated array object has register storage class. (I'm not sure why
it's undefined behavior rather than a constraint violation.)n

So is this code acceptable in C99?

#include <stdio.h>

int main(void)
{
register int a[2] = {
printf("%zu\n", sizeof *a),
printf("%zu\n", sizeof a)
};
return 0;
}

Since it's a normal array, and not a variable-length array, the
expression *a should never be evaluated, but just the type examined to
determine its size.

 

[Keith Thompson]

Keith Thompson wrote: [...]

6.5.4.2 says you can't apply unary "&" to a register-qualified
object.
6.3.2.1 says that the implicit conversion of an array expression to a
pointer to its first element invokes undefined behavior if the
designated array object has register storage class. (I'm not sure why
it's undefined behavior rather than a constraint violation.)n

So is this code acceptable in C99?

#include <stdio.h>

int main(void)
{
register int a[2] = {
printf("%zu\n", sizeof *a),
printf("%zu\n", sizeof a)
};
return 0;
}

Since it's a normal array, and not a variable-length array, the
expression *a should never be evaluated, but just the type examined to
determine its size.

Well, *I* certainly wouldn't accept it. }

But yes, I believe it's legal (no constraint violations, no undefined
behavior, but implementation-defined behavior because the output
depends on sizeof(int)).

<OT>
gcc complains "error: address of register variable 'a' requested" on
"sizeof *a". I believe gcc is wrong, but of course spurious
diagnostics don't make an implementation non-conforming.
</OT>

 

[Jordan Abel]

Keith Thompson wrote: [...]

6.5.4.2 says you can't apply unary "&" to a register-qualified
object.
6.3.2.1 says that the implicit conversion of an array expression to a
pointer to its first element invokes undefined behavior if the
designated array object has register storage class. (I'm not sure why
it's undefined behavior rather than a constraint violation.)n

So is this code acceptable in C99?

#include <stdio.h>

int main(void)
{
register int a[2] = {
printf("%zu\n", sizeof *a),
printf("%zu\n", sizeof a)
};
return 0;
}

Since it's a normal array, and not a variable-length array, the
expression *a should never be evaluated, but just the type examined to
determine its size.

Well, *I* certainly wouldn't accept it. }

But yes, I believe it's legal (no constraint violations, no undefined
behavior, but implementation-defined behavior because the output
depends on sizeof(int)).

<OT>
gcc complains "error: address of register variable 'a' requested" on
"sizeof *a". I believe gcc is wrong, but of course spurious
diagnostics don't make an implementation non-conforming.

I believe spurious _errors_ do. had it been a warning you'd be fine.
and are you sure it's not complaining at "sizeof a"? that's the one
where an address would be taken were the value to be used.

 

[Skarmander]

Keith Thompson wrote:
[...]

6.5.4.2 says you can't apply unary "&" to a register-qualified
object.
6.3.2.1 says that the implicit conversion of an array expression to a
pointer to its first element invokes undefined behavior if the
designated array object has register storage class. (I'm not sure why
it's undefined behavior rather than a constraint violation.)n

So is this code acceptable in C99?

#include <stdio.h>

int main(void)
{
register int a[2] = {
printf("%zu\n", sizeof *a),
printf("%zu\n", sizeof a)
};
return 0;
}

Since it's a normal array, and not a variable-length array, the
expression *a should never be evaluated, but just the type examined to
determine its size.

Well, *I* certainly wouldn't accept it. }

But yes, I believe it's legal (no constraint violations, no undefined
behavior, but implementation-defined behavior because the output
depends on sizeof(int)).

<OT>
gcc complains "error: address of register variable 'a' requested" on
"sizeof *a". I believe gcc is wrong, but of course spurious
diagnostics don't make an implementation non-conforming.
</OT>

I believe gcc is not wrong and that there *is* undefined behavior. I'm
not a language lawyer, so see if you can poke holes in this.

"Except when it is the operand of the sizeof operator or the unary &
operator, or is a string literal used to initialize an array, an
expression that has type "array of type" is converted to an expression
with type "pointer to type" that points to the initial element of the
array object and is not an lvalue. If the array object has register
storage class, the behavior is undefined."

Now *do not* tell me about the "except when it is the operand of the
sizeof operator" part. I *read it*. In the expression `sizeof *a', `a'
*is not* the operand of the sizeof operator -- `*a' is. Although `*a' is
not evaluated, its type must be determined. In order to do that, `a'
must be converted to a pointer (otherwise `*a' is a constraint
violation) and this conversion invokes undefined behavior.

Conversely, `sizeof a' is perfectly fine since here the exception
applies: `a' is the operand of the sizeof operator and the type is known.

S.

 

[Keith Thompson]

Keith Thompson wrote: [...]
6.5.4.2 says you can't apply unary "&" to a register-qualified
object.
6.3.2.1 says that the implicit conversion of an array expression to a
pointer to its first element invokes undefined behavior if the
designated array object has register storage class. (I'm not sure why
it's undefined behavior rather than a constraint violation.)n


So is this code acceptable in C99?

#include <stdio.h>

int main(void)
{
register int a[2] = {
printf("%zu\n", sizeof *a),
printf("%zu\n", sizeof a)
};
return 0;
}

Since it's a normal array, and not a variable-length array, the
expression *a should never be evaluated, but just the type examined to
determine its size.

Well, *I* certainly wouldn't accept it. }

But yes, I believe it's legal (no constraint violations, no undefined
behavior, but implementation-defined behavior because the output
depends on sizeof(int)).

<OT>
gcc complains "error: address of register variable 'a' requested" on
"sizeof *a". I believe gcc is wrong, but of course spurious
diagnostics don't make an implementation non-conforming.

I believe spurious _errors_ do. had it been a warning you'd be fine.

Oops, my mistake. I tried it on two different platforms, with two
different versions of gcc. On one (gcc 3.4.4 on Cygwin), it produced
a warning; on the other (gcc 4.0.2 on Solaris 9), it produced the same
message, but as an error rather than as a warning. I cut-and-pasted
the output from the latter without noticing that it was different.

and are you sure it's not complaining at "sizeof a"? that's the one
where an address would be taken were the value to be used.

Yes, I'm sure. In "sizeof a", a is an array expression, and it's not
converted to a pointer type because it's the operand of sizeof. In
"sizeof *a", a is the operand of the unary "*" operator, so it would
normally be converted to a pointer -- except that the operand of
"sizeof" isn't evaluated (unless it's a VLA). Both expressions are ok
because of the sizeof, but for different reasons: one because the
operand of sizeof isn't converted from array to pointer, the other
because any operand of sizeof is not evaluated.

And I checked the line number in the warning/error message.

Looks like a gcc bug (but a fairly minor one IMHO).

 

[Keith Thompson]

Keith Thompson wrote: [...]

6.5.4.2 says you can't apply unary "&" to a register-qualified
object.
6.3.2.1 says that the implicit conversion of an array expression to a
pointer to its first element invokes undefined behavior if the
designated array object has register storage class. (I'm not sure why
it's undefined behavior rather than a constraint violation.)n


So is this code acceptable in C99?

#include <stdio.h>

int main(void)
{
register int a[2] = {
printf("%zu\n", sizeof *a),
printf("%zu\n", sizeof a)
};
return 0;
}

Since it's a normal array, and not a variable-length array, the
expression *a should never be evaluated, but just the type examined to
determine its size.

Well, *I* certainly wouldn't accept it. }
But yes, I believe it's legal (no constraint violations, no undefined
behavior, but implementation-defined behavior because the output
depends on sizeof(int)).
<OT>
gcc complains "error: address of register variable 'a' requested" on
"sizeof *a". I believe gcc is wrong, but of course spurious
diagnostics don't make an implementation non-conforming.
</OT>

I believe gcc is not wrong and that there *is* undefined behavior. I'm
not a language lawyer, so see if you can poke holes in this.

"Except when it is the operand of the sizeof operator or the unary &
operator, or is a string literal used to initialize an array, an
expression that has type "array of type" is converted to an expression
with type "pointer to type" that points to the initial element of the
array object and is not an lvalue. If the array object has register
storage class, the behavior is undefined."

Now *do not* tell me about the "except when it is the operand of the
sizeof operator" part. I *read it*. In the expression `sizeof *a', `a'
*is not* the operand of the sizeof operator -- `*a' is. Although `*a'
is not evaluated, its type must be determined. In order to do that,
`a' must be converted to a pointer (otherwise `*a' is a constraint
violation) and this conversion invokes undefined behavior.

I don't think so. Any conversion would be part of the evaluation of
the expression. Because the expression "*a" as a whole is not
evaluated, the conversion doesn't take place. The compiler is (and
must be) perfectly capable of determining the type, and therefore the
size, of the expression without actually evaluating it.

Conversely, `sizeof a' is perfectly fine since here the exception
applies: `a' is the operand of the sizeof operator and the type is
known.

Agreed.

 

[Mark McIntyre]

On Thu, 10 Nov 2005 00:42:16 GMT, in comp.lang.c ,

I think we can assume that the OP meant "CPU register" when he said
"cache".

Ya reckon?

Sez you.

Yup, Says I. All the answers given are completely incorrect.

 

[Skarmander]

Keith Thompson wrote:

[...]


6.5.4.2 says you can't apply unary "&" to a register-qualified
object.
6.3.2.1 says that the implicit conversion of an array expression to a
pointer to its first element invokes undefined behavior if the
designated array object has register storage class. (I'm not sure why
it's undefined behavior rather than a constraint violation.)n


So is this code acceptable in C99?

#include <stdio.h>

int main(void)
{
register int a[2] = {
printf("%zu\n", sizeof *a),
printf("%zu\n", sizeof a)
};
return 0;
}

Since it's a normal array, and not a variable-length array, the
expression *a should never be evaluated, but just the type examined to
determine its size.

Well, *I* certainly wouldn't accept it. }
But yes, I believe it's legal (no constraint violations, no undefined
behavior, but implementation-defined behavior because the output
depends on sizeof(int)).
<OT>
gcc complains "error: address of register variable 'a' requested" on
"sizeof *a". I believe gcc is wrong, but of course spurious
diagnostics don't make an implementation non-conforming.
</OT>

I believe gcc is not wrong and that there *is* undefined behavior. I'm
not a language lawyer, so see if you can poke holes in this.

"Except when it is the operand of the sizeof operator or the unary &
operator, or is a string literal used to initialize an array, an
expression that has type "array of type" is converted to an expression
with type "pointer to type" that points to the initial element of the
array object and is not an lvalue. If the array object has register
storage class, the behavior is undefined."

Now *do not* tell me about the "except when it is the operand of the
sizeof operator" part. I *read it*. In the expression `sizeof *a', `a'
*is not* the operand of the sizeof operator -- `*a' is. Although `*a'
is not evaluated, its type must be determined. In order to do that,
`a' must be converted to a pointer (otherwise `*a' is a constraint
violation) and this conversion invokes undefined behavior.

I don't think so. Any conversion would be part of the evaluation of
the expression.

This I dispute. The standard does not mark conversion as belonging
exclusively to the evaluation process.

Because the expression "*a" as a whole is not evaluated, the conversion doesn't take place.

I agree that neither `a' nor `*a' are evaluated.

The compiler is (and must be) perfectly capable of determining the
type, and therefore the size, of the expression without actually
evaluating it.

I would be inclined to agree with you if you pointed out the rules in
the standard that specify what type `*a' has in this case, without
requiring any conversion.

The compiler can only determine the type according to the rules laid
down for this in the standard. The only rule I can find that specifies
what the type of `*a' is is given in 6.5.3.2.4: "[..] If the operand has
type 'pointer to type', the result has type 'type'." The constraints
specify that "the operand of the unary * operator shall have pointer
type", so that this type determination always applies.

But `a' is not of pointer type -- without conversion, `*a' is a
constraint violation. `a' is *not* the operand of a sizeof operator but
of an `*' operator, therefore it is converted to an `int*' per
6.3.2.1.3, and invokes undefined behavior per the same.

If you believe an alternate mechanism is required or supplied by the
standard that the compiler can apply to determine the type without going
through conversion, I'd like to know. I don't see what magically tells
the compiler what the type of `*a' is without tripping over the
conversion rules.

You can try and argue that the conversion is (part of) evaluation and
the compiler has only to *act* as if it converted for the sake of type
calculation, but must not actually do so. This is an eminently
reasonable approach, but it's made up out of whole cloth; the standard
doesn't say this is how determining types can be done. We are then
claiming the part about undefined behavior should be ignored without
having anything in the standard to back us up on this.

I repeat: I agree that neither `a' nor `*a' is actually evaluated, but I
do not agree that conversion could or should only happen on evaluation,
and I also argue that the conversion is necessary in this case. If this
is not the intent of the standard, I think it does not clearly specify
how the types of expressions are to be determined.

For contrast, consider `sizeof (0 / 0)'. The type of `0 / 0' can be
deduced by sections in the standard (namely 6.4.4.1, 6.3.1.8 and the
lack of any special provisions on the result type in 6.5.5) without
needing 6.5.5.5, which specifies UB for determining the result of a
division by zero. This is irrelevant because we do not evaluate the
expression and do not determine a result. Therefore `sizeof (0 / 0)' is
a well-defined expression equivalent to `sizeof int'. A similarly
careful analysis does not seem to give us any typing information for
`*a' that doesn't involve a clause with undefined behavior.

(We could take this to comp.std.c., it's getting rather technical.)

S.

 

[Keith Thompson]

This I dispute. The standard does not mark conversion as belonging
exclusively to the evaluation process.

I'm not sure I can cite anything that absolutely proves my point, but
here are a few quotes:

C99 6.3p1, "Conversions":

Several operators convert operand values from one type to another
automatically. This subclause specifies the result required from
such an _implicit conversion), as well as those that result from a
cast operation (an _explicit conversion_).

C99 6.5p1, "Expressions":

An _expression_ is a sequence of operators and operands that
specifies computation of a value, or that designates an object or
a function, or that generates side effects, or that performs a
combination thereof.

(This is a flawed definition, since there are expressions that contain
neither operators nor operands, but that's not relevant here.)

And, just for completeness:

C99 6.5.3.4p2, "The sizeof operator":

If the type of the operand is a variable length array type, the
operand is evaluated; otherwise, the operand is not evaluated and
the result is an integer constant.

It seems to me from these definitions that conversions are tied very
tightly to expressions. Conversions occur only as a result of an
operator being applied to an operand (not quite correct, but close
enough), which occurs as part of expression evaluation.

Furthermore, just as a matter of common sense (dangerous, I know),
expression evaluation happens at execution time; type determination
must happen at compilation time.

[snip]

If you believe an alternate mechanism is required or supplied by the
standard that the compiler can apply to determine the type without
going through conversion, I'd like to know. I don't see what magically
tells the compiler what the type of `*a' is without tripping over the
conversion rules.

There's nothing magical about it. For any non-VLA expression that's
the operand of a sizeof operator, the compiler has to go through the
same process it does for any expression to determine the type, but
without evaluating the expression (or rather, without generating code
to evaluate the expression).

You can try and argue that the conversion is (part of) evaluation and
the compiler has only to *act* as if it converted for the sake of type
calculation, but must not actually do so. This is an eminently
reasonable approach, but it's made up out of whole cloth; the standard
doesn't say this is how determining types can be done. We are then
claiming the part about undefined behavior should be ignored without
having anything in the standard to back us up on this.

Then how *does* the standard say the type is determined? If it has to
carry out a run-time operation (the conversion) to determine
compile-time information, we're in big trouble.