Malloc

[pavan]

Is this code correct

char *str = (char *)malloc(10 * sizeof(char));
char[0] = ...
char [1] = ...
....
....
...
char [9] = '\0';

malloc() does not make any guarantees regarding the contiguity of
memory. In that case, is ths code correct? Can we assume that the
characters are contiguous in memory.

[Ico]

Re: Malloc

pavan <pavan.mail@gma il.comwrote:

malloc() does not make any guarantees regarding the contiguity of
memory.

Why do you think this is the case ?

--
:wq
^X^Cy^K^X^C^C^C ^C

[pavan]

Re: Malloc

On Feb 3, 11:15 am, Ico <use...@zeev.nl wrote:

pavan <pavan.m...@gma il.comwrote:

malloc() does not make any guarantees regarding the contiguity of
memory.

>
Why do you think this is the case ?
>
--
:wq
^X^Cy^K^X^C^C^C ^C

if i create a chunk of memory containing a number of elements of the
same type, i want to know if i can access the elements as ptr, ptr+1,
ptr+2 etc... Should I use vmalloc() instead of malloc if i want to do
this?

[santosh]

Re: Malloc

pavan wrote:

On Feb 3, 11:15 am, Ico <use...@zeev.nl wrote:

pavan <pavan.m...@gma il.comwrote:

malloc() does not make any guarantees regarding the contiguity of
memory.

Why do you think this is the case ?

>
if i create a chunk of memory containing a number of elements of the
same type, i want to know if i can access the elements as ptr, ptr+1,
ptr+2 etc...

Yes, you can.

Should I use vmalloc() instead of malloc if i want to do this?

No. vmalloc() is a non-standard function. malloc() is fine.

[pavan]

Re: Malloc

Does this mean that malloc returns memory that is contiguous in
virtual address space?

[santosh]

Re: Malloc

pavan wrote:

Is this code correct
>
char *str = (char *)malloc(10 * sizeof(char));
char[0] = ...
char [1] = ...

I hope you mean str[0], str[1] etc.? char is a type. Types are not
directly usable. You must have an object of a type, to use it.

...
...
..
char [9] = '\0';
>
malloc() does not make any guarantees regarding the contiguity of
memory.

Where do get this nonsense from. A chunk of memory returned by
malloc() calloc() or realloc() can be regarded as an array, which is
by definition contiguous.

In that case, is ths code correct? Can we assume that the
characters are contiguous in memory.

Yes. Elements in a block allocated by a single call to *alloc() can be
regarded as sequential.

[santosh]

Re: Malloc

pavan wrote:

Does this mean that malloc returns memory that is contiguous in
virtual address space?

>From the point of view of a C program, it is contiguous. Whether it's

so in virtual/physical/whatever memory is outside the scope of C and
moreover, totally irrelevant to most C programs.

[Clark S. Cox III]

Re: Malloc

pavan wrote:

Is this code correct
>
char *str = (char *)malloc(10 * sizeof(char));

Don't cast the result of malloc. The sizeof(char) is redundant, as
sizeof(char) is always 1.

char[0] = ...
char [1] = ...
....
....
...
char [9] = '\0';
>
malloc() does not make any guarantees regarding the contiguity of
memory.

Of course it does, who told you otherwise?

In that case, is ths code correct? Can we assume that the
characters are contiguous in memory.
>

--
Clark S. Cox III
clarkcox3@gmail .com

[Clark S. Cox III]

Re: Malloc

pavan wrote:

Does this mean that malloc returns memory that is contiguous in
virtual address space?
>

As far as any conforming C program is concerned, malloc returns
contiguous memory. Period.

--
Clark S. Cox III
clarkcox3@gmail .com

[Mark McIntyre]

Re: Malloc

On 3 Feb 2007 08:29:50 -0800, in comp.lang.c , "pavan"
<pavan.mail@gma il.comwrote:

>Does this mean that malloc returns memory that is contiguous in
>virtual address space?

The C language doesn't need to know anything about how the OS lays out
memory in any virtual address space. As far as C is concerned, any
memory returned by malloc is a contiguous block.

If your programme definitely needs to know precisely how your OS is
choosing and allocating memory, it may be that C is the wrong language
to use.


--
Mark McIntyre

"Debugging is twice as hard as writing the code in the first place.
Therefore, if you write the code as cleverly as possible, you are,
by definition, not smart enough to debug it."
--Brian Kernighan

[Richard Heathfield]

Re: Malloc

pavan said:

Is this code correct
>
char *str = (char *)malloc(10 * sizeof(char));
char[0] = ...
char [1] = ...
...
...
..
char [9] = '\0';

No. Wrap it in a function, lose the cast, add a check to determine whether
str is NULL after the malloc call, change sizeof(char) to sizeof *str,
#include <stdlib.hand change char[0] etc to str[0] etc, and you'll have
something you can use.

malloc() does not make any guarantees regarding the contiguity of
memory.

"The order and contiguity of storage allocated *by successive calls*
to the calloc , malloc , and realloc functions is unspecified." Since malloc
can be used to allocate storage for any object, we may deduce that the
storage thus allocated in a single call is contiguous.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999

http://www.cpax.org.uk

email: rjh at the above domain, - www.

[santosh]

Re: Malloc

Mark McIntyre wrote:

On 3 Feb 2007 08:29:50 -0800, in comp.lang.c , "pavan"
<pavan.mail@gma il.comwrote:
>

Does this mean that malloc returns memory that is contiguous in
virtual address space?

>
The C language doesn't need to know anything about how the OS lays out
memory in any virtual address space. As far as C is concerned, any
memory returned by malloc is a contiguous block.
>
If your programme definitely needs to know precisely how your OS is
choosing and allocating memory, it may be that C is the wrong language
to use.

Why? Most of the OSes are written in C.

[jacob.navia]

Re: Malloc

pavan a écrit :

Does this mean that malloc returns memory that is contiguous in
virtual address space?
>

Yes

[CBFalconer]

Re: Malloc

pavan wrote:

>
Is this code correct
>
char *str = (char *)malloc(10 * sizeof(char));
char[0] = ...
char [1] = ...
...
...
..
char [9] = '\0';
>
malloc() does not make any guarantees regarding the contiguity of
memory. In that case, is ths code correct? Can we assume that the
characters are contiguous in memory.

Yes you can, but you should correct your call on malloc.
sizeof(char) is 1 by definition, and the only function of the cast
is to suppress the needed error message which will point out the
failure to #include <stdlib.h>. A better phrase would be:

char *str = malloc(10 * sizeof *str);

which can be retyped by changing only the "char", and will not
suppress any error messages.

--
<http://www.cs.auckland .ac.nz/~pgut001/pubs/vista_cost.txt>

"A man who is right every time is not likely to do very much."
-- Francis Crick, co-discover of DNA
"There is nothing more amazing than stupidity in action."
-- Thomas Matthews

[Kenneth Brody]

Re: Malloc

santosh wrote:

>
Mark McIntyre wrote:

[...]

If your programme definitely needs to know precisely how your OS is
choosing and allocating memory, it may be that C is the wrong language
to use.

>
Why? Most of the OSes are written in C.

In clc, "the C language" means "as defined by the ISO standard".
More precisely, what he probably meant was "you will need to go
outside the scope of the standard, and use some system- specific
methods to do what you need". Whether that means using C for most
of your work, plus using some implementation-specific additions,
or using a different language entirely, is up to you.

That, and I have no idea what percentage of operating systems are
written in C. However, given clc's definition of C, I would
venture that, as far as clc is concerned, no true "operating
system" is, or possibly even could be, written in standard C.

--
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