Does this allocate memory?

[Curt]

If this is the complete program (ie, the address of the const is never
taken, only its value used) is it likely the compiler will allocate ram for
constantA or constantB? Or simply substitute the values in (as would be
required if I used the hideous, evil, much-abused #define :)

-----------

const int constantA = 10;
static const int constantB = 20;

void main()
{
for( int i=0; i<constantA ; i++ );
for( int j=0; j<constantB ; j++ );
}

----------

Thanks in advance :)

-Curt

[Alf P. Steinbach]

On Tue, 22 Jul 2003 13:49:30 GMT, Curt <cN****@nSoPrth AarMc.com> wrote:

If this is the complete program (ie, the address of the const is never
taken, only its value used) is it likely the compiler will allocate ram for
constantA or constantB? Or simply substitute the values in (as would be
required if I used the hideous, evil, much-abused #define :)

-----------

const int constantA = 10;
static const int constantB = 20;

void main()
{
for( int i=0; i<constantA ; i++ );
for( int j=0; j<constantB ; j++ );
}

A good compiler will reject this program since it's ill-formed;
"void main()" is not allowed in a hosted standard C++
implemented (the term "hosted" is the standard's terminology).

A not-so-good compiler will reduce the program to nothing since
it doesn't do anything.

But in general, any constant takes up memory space if it's used
at least once. How much is Quality Of Implementation issue.
Not much more can be said without a more specific context.

[Victor Bazarov]

"Curt" <cN****@nSoPrth AarMc.com> wrote...

If this is the complete program (ie, the address of the const is never
taken, only its value used) is it likely the compiler will allocate ram for constantA or constantB? Or simply substitute the values in (as would be
required if I used the hideous, evil, much-abused #define :)

-----------

const int constantA = 10;
static const int constantB = 20;

void main()
{
for( int i=0; i<constantA ; i++ );
for( int j=0; j<constantB ; j++ );
}

----------

It's hard to tell. Existence of 'void main' can force the compiler
reject the entire program. 'main' always returns 'int'. If that's
corrected, an optimising compiler will produce code as if the program
were

int main() {}

because the body of the 'main' has no side effects. So, once the
'main' is made to return 'int', I'd say, no memory is going to be
allocated.

Victor

[Curt]

"Victor Bazarov" <v.********@att Abi.com> wrote in news:vhqgt0eifs 2499
@corp.supernews .com:

"Curt" <cN****@nSoPrth AarMc.com> wrote...

If this is the complete program (ie, the address of the const is never
taken, only its value used) is it likely the compiler will allocate ram

for

constantA or constantB? Or simply substitute the values in (as would

be required if I used the hideous, evil, much-abused #define :)

-----------

const int constantA = 10;
static const int constantB = 20;

void main()
{
for( int i=0; i<constantA ; i++ );
for( int j=0; j<constantB ; j++ );
}

----------

It's hard to tell. Existence of 'void main' can force the compiler
reject the entire program. 'main' always returns 'int'. If that's
corrected, an optimising compiler will produce code as if the program
were

int main() {}

because the body of the 'main' has no side effects. So, once the
'main' is made to return 'int', I'd say, no memory is going to be
allocated.

Victor

I'll assume you're not missing my point on purpose, which is, does the
'static' qualifier compel the compiler to allocate memory for a constant
that is initialized once at runtime and is only used as a #define might
be.
const int constantA = 10;
static const int constantB = 20;

int main( int argn, char *argv[] )
{
for( volatile int i=0; i<constantA ; i++ );
for( volatile int j=0; j<constantB ; j++ );

return 0;
}

[Peter van Merkerk]

> >> const int constantA = 10;

static const int constantB = 20;

void main()
{
for( int i=0; i<constantA ; i++ );
for( int j=0; j<constantB ; j++ );
}

----------
It's hard to tell. Existence of 'void main' can force the compiler
reject the entire program. 'main' always returns 'int'. If that's
corrected, an optimising compiler will produce code as if the program were

int main() {}

because the body of the 'main' has no side effects. So, once the
'main' is made to return 'int', I'd say, no memory is going to be
allocated.

I'll assume you're not missing my point on purpose, which is, does the
'static' qualifier compel the compiler to allocate memory for a constant that is initialized once at runtime and is only used as a #define might be.
The 'static' keyword does not force the compiler to allocate storage for
that variable. In fact it would make it easier for the compiler not to
allocate storage since static variables only accessible within the
current translation unit. Consequently the compiler can determine with
100% certainty that a variable is not referenced, thus no storage needs
to be allocated for it. Storage for non 'static' variables would have to
be removed by the linker. The fact that the variables in your example
are declared 'const' and that they are initialized in the same
translation unit makes it likely that a optimizing compiler simply
substitutes the variables with their values. In that scenario the
variables in your code are not referenced, thus no storage is needed for
those variables. So my guess is that on most decent optimizing compilers
your program will be optimized to int main(){} without storage for the
variables.

However all of this is a quality of implementation issue and your
compiler&linker may, or may not perform those optimizations. If you
really want to know if which optimizations are performed take a look at
the assembly output of the compiler. Note that this output doesn't tell
you which variables were removed by the linker, so I expect that
constantA still has storage associated with it according to listing,
even though it might very well be removed in the linking stage.
const int constantA = 10;
static const int constantB = 20;

int main( int argn, char *argv[] )
{
for( volatile int i=0; i<constantA ; i++ );
for( volatile int j=0; j<constantB ; j++ );

return 0;
}

Why are you using volatile for a local variable?

--
Peter van Merkerk
peter.van.merke rk(at)dse.nl

[Victor Bazarov]

"Curt" <cN****@nSoPrth AarMc.com> wrote...

[...]
I'll assume you're not missing my point on purpose, which is, does the
'static' qualifier compel the compiler to allocate memory for a constant
that is initialized once at runtime and is only used as a #define might
be.

'static' has no effect on the constant. Used with a variable
definition in a global namespace scope, 'static' give it internal
linkage. However, if memory is not allocated, linkage would have
no meaning. 'const' allows the compiler not to allocate memory.
That's the prevailing modifier, I believe. 'static' in this case
is subordinate.

Just as with any other 'const', the compiler may decide to allocate
storage. Or it may decide not to allocate storage.

If you wrote

extern const int theanswer = 42;

then the storage would be allocated for sure. Whether the address
of that storage will be used when you use 'theanswer' in the same
unit is also at the discretion of the compiler. It may decide to
replace any occurrence of 'theanswer' in the code with 42 because
it's a 'const'.

Does this answer your question?

Victor

[Jingz]

[...]

Thank you for your response.

const int constantA = 10;
static const int constantB = 20;

int main( int argn, char *argv[] )
{
for( volatile int i=0; i<constantA ; i++ ); for( volatile int j=0;
j<constantB ; j++ );

return 0;
}
}

Why are you using volatile for a local variable?

Because the original responders wanted to tell me how smart they were
about proper c++ rather than answer the obvious question in a meaningful
way. So I composed an equally trivial example that would force code
generation.

As a long-time programmer I am sometimes amused (and dismayed) that such
simple language features ( #define constant 10 ) are universally derided
by acedemia/language purists who insist on type-safe const's, yet are not
sure if it will have the same necessary effect of being replaced with the
literal value.

c++ took the language in a lot of good directions, but went too far in
others, the useless defense of cout vs printf for debugging, for example.

-Curt

[Ron Natalie]

"Jingz" <[e-mail]> wrote in message news:pa******** *************** *********@spam. com...

As a long-time programmer I am sometimes amused (and dismayed) that such
simple language features ( #define constant 10 ) are universally derided
by acedemia/language purists who insist on type-safe const's, yet are not
sure if it will have the same necessary effect of being replaced with the
literal value.

It's not typesafety. 10 is of type int just as if you declared it a const int.
The issue is scoping. What happens when something else in the translation
unit (perhaps something you didn't write) use the term constant:

#define constant 10

struct foo {
int constant;
};
....

Perhaps it's just that you don't understand the language well enough.

[Jingz]

On Tue, 22 Jul 2003 14:52:59 -0400, Ron Natalie wrote:

"Jingz" <[e-mail]> wrote in message
news:pa******** *************** *********@spam. com...

As a long-time programmer I am sometimes amused (and dismayed) that
such simple language features ( #define constant 10 ) are universally
derided by acedemia/language purists who insist on type-safe const's,
yet are not sure if it will have the same necessary effect of being
replaced with the literal value.

It's not typesafety. 10 is of type int just as if you declared it a
const int. The issue is scoping. What happens when something else in
the translation unit (perhaps something you didn't write) use the term
constant:

#define constant 10

struct foo {
int constant;
};
...

Perhaps it's just that you don't understand the language well enough.

Yes I've seen that argument as well, it is equally contrived. You really
have to bend over backwards to show how sane use of #define can be a
problem; your example doesn't even compile. A trivial convention like
using all capital letters for #defined values is all that would be
required.

I'm not going to defend such a practice, nor expouse it. The point I have
often made when instructing junior programmers fresh out of college is
that many textbook problems are solutions to problems that only exist in
textbooks.

"The real world" is an overused phrase, but maintainability is key. Too
often I have read in a trade-journal or commentary about a perfectly
reasonable practice that is "bad' becuase some college professor can
contrive an example that "breaks" is.

Not to open up another can of worms, but inheritance is probobly my
favorite example of a good idea gone bad. Everyone agree is it can be
over/mis used, but I will content that it is almost never a good idea. It
obscures functionality at best, at worst is is an absolutely impenatrable
series of tracing back multiple-inheritance spaghetti when a call goes
bad and needs to be debugged. 'has a' is far superior, necessitating a
dereference and obviating that another block of code is being invoked.
"is a" is almost never justified.

Sure it allows library building, and has a nice touchy-feely OO, but the
fact is it solves a bunch of contrived textbook problems that proper
program design not only bypass, but make easier to understand.

-Curt

[Jingz]

I know I can't win, you are in quite authoritative company in terms of
people I've argued with about programming paradigms, and I know anyone
reading this agrees with you, not me, but I do feel compelled to stick to
facts.

Microsoft thinks defined like "BOOL" and "WORD" make sense, so I'm not
quite sure invoking problems with their source code is credible. I have
indeed never seen a sane define collide in the manner you suggest happens
"all the time" but if we're going to assume insane programmers then you
can prety much claim anything you choose.

I certainly agree that c++ promotes code cloarity and maintainability , no
question about it, but te fact that it can is often used as a cudgel to
beat code with. Complicated multiple inheritance, templates, and bizzare operator
overloading are automatically easier to maintain? I think not. They CAN
be, but care must be used, as in all programming.

-Curt
On Tue, 22 Jul 2003 16:18:48 -0400, Ron Natalie wrote:

"Jingz" <[e-mail]> wrote in message
news:pa******** *************** ***********@spa m.com...

Yes I've seen that argument as well, it is equally contrived.

It's not contrived at all. If you've never seen it, you've never
managed a large project where more than one organization wrote parts of
the code.

You really
have to bend over backwards to show how sane use of #define can be a
problem;

I didn't bend over backwards. This stuff happens all the time. It's
a perennial problem with Microsoft include files for example.

I'm not going to defend such a practice, nor expouse it. The point I
have often made when instructing junior programmers fresh out of
college is that many textbook problems are solutions to problems that
only exist in textbooks.

Sorry, just because YOU have never seen the problem, doesn't mean they
shouldn't exist. Do you advocate telling drivers to not fasten their
seatbelts because you've never been in an accident?

"The real world" is an overused phrase, but maintainability is key.

Precisely. Using C++ constructs promotes maintainability .