how much memory is allocated for following structure
struct bharath
{
int b;
char c;
float d;
}
and how?
Jun 21 '07
43  2387 
Army1987 wrote:
>
.... snip ...
>
In something like
int main(void) {
struct bharath b;
long long ll;
...
}
since the standard requires nothing about the placement of auto
variables, it is well possible that there is some space between
b and ll, but an auto char in an inner block might well be put into
that space. (Yes, I don't think it is very likely to happen...)
So, as far as a reasonable definition of "to allocate" is
concerned, these two declarations allocate exactly sizeof b +
sizeof ll bytes, even if they are not contiguous.
A long long requires <sizeof long long\bytes, not 11.
--
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cbfalconer at maineline dot net
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Posted via a free Usenet account from http://www.teranews.com
Keith Thompson wrote:
Barry Schwarz <sc******@doezl .netwrites:
>On Sun, 24 Jun 2007 00:10:45 -0000, SM Ryan
<wy*****@tan go-sierra-oscar-foxtrot-tango.fake.orgw rote:
[...]
>>> struct bharath
{
int b;
char c;
float d;
}
A variable of type (struct bharath) will be allocated at
least sizeof(struct bharath) bytes. How a compiler allocates
Would you care to give an example of when it will be allocated more
than sizeof(srtuct bharath) bytes?
malloc(sizeof(s truct bharath)) can easily allocate more than
sizeof(struct bharath) bytes. Even for an object declaration, there
may be a gap between the object and the next object in memory.
I don't think so. First, sizeof (struct bharath) is 12 at my house. The
offsetof the members are 0, 4 and 8 regardless of their arrangement. If
the char member is the last it is still at offset 8 and sizeof the
struct is still 12. There can be no gap between the object and the next
object.
You could argue, of course, that the extra bytes aren't allocated *to
the object*, but are just padding outside the object (and I wouldn't
disagree).
There are no 'extra bytes' that you can find. That malloc might allocate
more than asked for is true but not at issue.
struct bharath *bhar = malloc(3 * sizeof *bhar);
If successful, this will allocate at least 36 bytes for three instances
of the struct. malloc may allocate more for its own purposes but we
can't know and don't care.
Or did I misunderstand you?
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
CBFalconer wrote:
Army1987 wrote:
>>
... snip ...
>>
In something like
int main(void) {
struct bharath b;
long long ll;
...
}
since the standard requires nothing about the placement of auto
variables, it is well possible that there is some space between
b and ll, but an auto char in an inner block might well be put into
that space. (Yes, I don't think it is very likely to happen...)
So, as far as a reasonable definition of "to allocate" is
concerned, these two declarations allocate exactly sizeof b +
sizeof ll bytes, even if they are not contiguous.
A long long requires <sizeof long long\bytes, not 11.
A variable named LL requires sizeof LL bytes, and the lowercase equivalent
requires sizeof ll bytes.
CBFalconer <cb********@yah oo.comwrites:
Army1987 wrote:
>>
... snip ...
>>
In something like
int main(void) {
struct bharath b;
long long ll;
...
}
since the standard requires nothing about the placement of auto
variables, it is well possible that there is some space between
b and ll, but an auto char in an inner block might well be put into
that space. (Yes, I don't think it is very likely to happen...)
So, as far as a reasonable definition of "to allocate" is
concerned, these two declarations allocate exactly sizeof b +
sizeof ll bytes, even if they are not contiguous.
A long long requires <sizeof long long\bytes, not 11.
Chuck, if you can't tell the difference between "l" (lowercase L) and
"1" (digit one), get a better font. At least take a closer look
before assuming that somebody wrote something nonsensical like "sizeof
11". (Yes, "sizeof 11" is legal, but it makes no sense in context.)
--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Joe Wright <jo********@com cast.netwrites:
Keith Thompson wrote:
>Barry Schwarz <sc******@doezl .netwrites:
>>On Sun, 24 Jun 2007 00:10:45 -0000, SM Ryan
<wy*****@tang o-sierra-oscar-foxtrot-tango.fake.orgw rote:
[...]
>>>> struct bharath
{
int b;
char c;
float d;
}
A variable of type (struct bharath) will be allocated at
least sizeof(struct bharath) bytes. How a compiler allocates
Would you care to give an example of when it will be allocated more
than sizeof(srtuct bharath) bytes?
malloc(sizeof( struct bharath)) can easily allocate more than
sizeof(struc t bharath) bytes. Even for an object declaration, there
may be a gap between the object and the next object in memory.
I don't think so. First, sizeof (struct bharath) is 12 at my
house. The offsetof the members are 0, 4 and 8 regardless of their
arrangement. If the char member is the last it is still at offset 8
and sizeof the struct is still 12. There can be no gap between the
object and the next object.
Sure there can.
Suppose sizeof(struct bharath) is 12, and suppose you have another
type struct foo that requires 16-byte alignment. Then if you declare
struct bharath obj1;
struct foo obj2;
there easily *could* be a 4-byte gap between obj1 and obj2. Or the
compiler could reverse the order, or it could allocate another object
between obj1 and obj2, or whatever.
I'm not saying this is of any particular importance, just that it's
possible and plausible.
I'm sure you know this, and I suspect I'm misinterpreting your
statement that "There can be no gap between the object and the next
object."; can you clarify what you mean?
>You could argue, of course, that the extra bytes aren't allocated *to
the object*, but are just padding outside the object (and I wouldn't
disagree).
There are no 'extra bytes' that you can find. That malloc might
allocate more than asked for is true but not at issue.
Ok, so what exactly is the issue? I think I've lost track.
I think we all know what the answers are, and we're arguing over what
the question is, or perhaps what the question should be.
[...]
--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Keith Thompson wrote:
Joe Wright <jo********@com cast.netwrites:
>Keith Thompson wrote:
>>Barry Schwarz <sc******@doezl .netwrites:
On Sun, 24 Jun 2007 00:10:45 -0000, SM Ryan
<wy*****@tan go-sierra-oscar-foxtrot-tango.fake.orgw rote:
[...]
> struct bharath
> {
> int b;
> char c;
> float d;
> }
A variable of type (struct bharath) will be allocated at
least sizeof(struct bharath) bytes. How a compiler allocates
Would you care to give an example of when it will be allocated more
than sizeof(srtuct bharath) bytes?
malloc(sizeof (struct bharath)) can easily allocate more than
sizeof(stru ct bharath) bytes. Even for an object declaration, there
may be a gap between the object and the next object in memory.
I don't think so. First, sizeof (struct bharath) is 12 at my
house. The offsetof the members are 0, 4 and 8 regardless of their
arrangement. If the char member is the last it is still at offset 8
and sizeof the struct is still 12. There can be no gap between the
object and the next object.
Sure there can.
Suppose sizeof(struct bharath) is 12, and suppose you have another
type struct foo that requires 16-byte alignment. Then if you declare
struct bharath obj1;
struct foo obj2;
there easily *could* be a 4-byte gap between obj1 and obj2. Or the
compiler could reverse the order, or it could allocate another object
between obj1 and obj2, or whatever.
I wasn't thinking about a gap between obj1 and obj2. If there is such a
gap we don't care and I think we can't even test for it.
I'm not saying this is of any particular importance, just that it's
possible and plausible.
I'm sure you know this, and I suspect I'm misinterpreting your
statement that "There can be no gap between the object and the next
object."; can you clarify what you mean?
I was alluding to an array of the particular structure.
>>You could argue, of course, that the extra bytes aren't allocated *to
the object*, but are just padding outside the object (and I wouldn't
disagree).
There are no 'extra bytes' that you can find. That malloc might
allocate more than asked for is true but not at issue.
Ok, so what exactly is the issue? I think I've lost track.
I think we all know what the answers are, and we're arguing over what
the question is, or perhaps what the question should be.
Indeed. I got sideways on the 'extra bytes' thing.
--
Joe Wright
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---
# >>Would you care to give an example of when it will be allocated more
# >>than sizeof(srtuct bharath) bytes?
# >malloc(sizeof( struct bharath)) can easily allocate more than
# >sizeof(struc t bharath) bytes. Even for an object declaration, there
# >may be a gap between the object and the next object in memory.
Sometimes a compiler will allocate variables in word sized units
to improve performance. As far as such compilers are concerned,
the variable is a whole number of words which can be greater
than the sizeof.
--
SM Ryan http://www.rawbw.com/~wyrmwif/
OOOOOOOOOO! NAVY SEALS!
On Jun 21, 10:33 am, Richard Heathfield <r...@see.sig.i nvalidwrote:
bharath...@gmai l.com said:
how much memory is allocated for following structure
struct bharath
{
int b;
char c;
float d;
}
None. It's a type, not an object.
A newbie question: surely, the struct definition is stored somewhere.
Else, how would the struct be allocated at runtime? And this would
require some memory...
Regards,
Frodo B
On Mon, 25 Jun 2007 01:21:58 -0000, SM Ryan
<wy*****@tang o-sierra-oscar-foxtrot-tango.fake.orgw rote:
># >>Would you care to give an example of when it will be allocated more
# >>than sizeof(srtuct bharath) bytes?
# >malloc(sizeof( struct bharath)) can easily allocate more than
# >sizeof(struc t bharath) bytes. Even for an object declaration, there
# >may be a gap between the object and the next object in memory.
Sometimes a compiler will allocate variables in word sized units
to improve performance. As far as such compilers are concerned,
the variable is a whole number of words which can be greater
than the sizeof.
No it cannot. The reason is that elements of an array must abut and
the byte difference between the start of one element and the next
element must be exactly sizeof element. If the compiler elects to pad
a structure out to a word boundary, then that padding is included the
value that sizeof evaluates to.
Remove del for email
Frodo Baggins said:
On Jun 21, 10:33 am, Richard Heathfield <r...@see.sig.i nvalidwrote:
>bharath...@gma il.com said:
how much memory is allocated for following structure
struct bharath
{
int b;
char c;
float d;
}
None. It's a type, not an object.
A newbie question: surely, the struct definition is stored somewhere.
Yes, the compiler will need to store the definition in its own internal
memory, but in those terms the question is unanswerable without
detailed internal knowledge of the compiler working. But that's not
normally what we mean when we say "how much memory".
Else, how would the struct be allocated at runtime?
It wouldn't be. It's a type, not an object. You don't allocate types.
--
Richard Heathfield <http://www.cpax.org.uk >
Email: -www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
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