memcat fn

A

aarklon

Hi all,

is it possible to write a function named memcat, which offers
functionality similar to that of the strcat fn, i.e i mean a function
on the following lines:-

void * memcat(void *s1, void *s2);

now s1 should point to the beginning of the concatenated memory
region. now as in strcat how to determine the terminating memory
location...???
 
M

Mark Bluemel

Hi all,

is it possible to write a function named memcat, which offers
functionality similar to that of the strcat fn, i.e i mean a function
on the following lines:-

void * memcat(void *s1, void *s2);

You need to define the functionality you expect from such a function.
now s1 should point to the beginning of the concatenated memory
region. now as in strcat how to determine the terminating memory
location...???

That, as the man said so memorably, is the question.
 
E

Eric Sosman

Hi all,

is it possible to write a function named memcat, which offers
functionality similar to that of the strcat fn, i.e i mean a function
on the following lines:-

void * memcat(void *s1, void *s2);

now s1 should point to the beginning of the concatenated memory
region. now as in strcat how to determine the terminating memory
location...???

Let's begin with the last question: How to determine
the end. Two ends, really, because you need to find the
end of the existing piece that begins at s1 and the end
of the added piece that begins at s2. The other memxxx
function (memset, memchr, ...) use byte counts for this:
the caller provides an extra argument giving the number
of bytes in the memory area. For memcat there are two
memory areas, hence two counts, and the function looks like

void *memcat(void *s1, size_t n1, void *s2, size_t n2);

Let's not stop there, though. Think for a minute about
what memcat will do, internally. All the area 1 bytes will
remain as they are, untouched, and the new material will be
added right after them. So memcat probably begins with

void *target = (char*)s1 + n1;

.... to get a pointer the the spot where the new material
will go. (The (char*) cast is needed because you can't do
arithmetic on a void* pointer.) What next? The rest of the
job is just copying the area 2 material to its new position.
So the complete implementation of memcat might look like

void *memcat(void *s1, size_t n1, void *s2, size_t n2) {
void *target = (char*)s1 + n1;
memcpy(target, s2, n2);
return s1;
}

.... or, with some abbreviation
void *memcat(void *s1, size_t n1, void *s2, size_t n2) {
memcpy((char*)s1 + n1, s2, n2);
return s1;
}

(An "industrial-strength" version would probably use const
on s2, and a version for C99 compilers would also use the
restrict qualifier, but the outline would be the same.)

In other words, memcat is just memcpy with a different
starting point! And that's probably why it doesn't exist in
the Standard library: it's a trivial variation on a function
that's already provided. Adding it would be a little bit like
adding a sqrt_half function that computed the square root of
one-half its argument: A task that's easily done by calling
the usual sqrt function with a halved argument to begin with.
 
M

Martien Verbruggen

Hi all,

is it possible to write a function named memcat, which offers
functionality similar to that of the strcat fn, i.e i mean a function
on the following lines:-

void * memcat(void *s1, void *s2);

now s1 should point to the beginning of the concatenated memory
region. now as in strcat how to determine the terminating memory
location...???
[snip]

void *memcat(void *s1, size_t n1, void *s2, size_t n2) {
memcpy((char*)s1 + n1, s2, n2);
return s1;
}

of course, if you wrote it like this, you'd have to include string.h, in
which case you shouldn't be naming your function memcat()i as long as it
has external linkage, as that would be a reserved identifier.

Martien
 


$)CHarald van D)&k

of course, if you wrote it like this, you'd have to include string.h, in
which case you shouldn't be naming your function memcat()i as long as it
has external linkage, as that would be a reserved identifier.

memcat is a reserved external name even if you don't include string.h.
Including string.h means that you're also not allowed to define static
void *memcat(...), or typedef void memcat, since it becomes reserved as a
file scope identifier.
 
C

CBFalconer

is it possible to write a function named memcat, which offers
functionality similar to that of the strcat fn, i.e i mean a
function on the following lines:-

void * memcat(void *s1, void *s2);

now s1 should point to the beginning of the concatenated memory
region. now as in strcat how to determine the terminating memory
location...???

Strings are terminated by the final '\0'. Memory blocks have no
such delimiter. Thus you have to have, and supply, the sizes of
the memory blocks. Therefore the prototype could be:

int memcat(char *s1, size_t sz1, /* input block & destination
*/
char *s2, size_t sz2, /* block to add to it */
size_t szout); /* maximum size of *s1 */

and it can return an error indicator, such as non-zero for failure
to fit everything.
 
A

aarklon

Hi all,
is it possible to write a function named memcat, which offers
functionality similar to that of the strcat fn, i.e i mean a function
on the following lines:-
void * memcat(void *s1, void *s2);
now s1 should point to the beginning of the concatenated memory
region. now as in strcat how to determine the terminating memory
location...???
[snip]



void *memcat(void *s1, size_t n1, void *s2, size_t n2) {
memcpy((char*)s1 + n1, s2, n2);
return s1;
}

of course, if you wrote it like this, you'd have to include string.h, in
which case you shouldn't be naming your function memcat()i as long as it
has external linkage, as that would be a reserved identifier.

Martien
--
|
Martien Verbruggen | Computers in the future may weigh no more
| than 1.5 tons. -- Popular Mechanics, 1949
|

I searched my /usr/include/string.h file for memcat but i didn't find
any. I use gcc v 4.01
 
F

Flash Gordon

On Nov 23, 4:43 pm, Martien Verbruggen <[email protected]>
wrote:

Please don't quote peoples signatures, the bit typically after the "-- "
or anything else not relevant to your reply.
I searched my /usr/include/string.h file for memcat but i didn't find
any. I use gcc v 4.01

Martien did not say it was defined, he said it was reserved. I.e. you
are *not* allowed to use that name whether it is currently used or not.
The GNU people could decide to use it in the next minor release thus
breaking your code, or they might be using some "compiler magic" which
well break your code, or they could perfectly legally just detect that
you have used it and deliberately generate code that generates random
insults.
 
B

Barry Schwarz

On Nov 23, 4:43 pm, Martien Verbruggen <[email protected]>
wrote:
snip


I searched my /usr/include/string.h file for memcat but i didn't find
any. I use gcc v 4.01

Which only proves that your system doesn't have it.

The real issue is that the name is reserved, regardless of whether it
is in use at the moment or not.


Remove del for email
 
A

aarklon

Let's begin with the last question: How to determine
the end. Two ends, really, because you need to find the
end of the existing piece that begins at s1 and the end
of the added piece that begins at s2. The other memxxx
function (memset, memchr, ...) use byte counts for this:
the caller provides an extra argument giving the number
of bytes in the memory area. For memcat there are two
memory areas, hence two counts, and the function looks like

void *memcat(void *s1, size_t n1, void *s2, size_t n2);

Let's not stop there, though. Think for a minute about
what memcat will do, internally. All the area 1 bytes will
remain as they are, untouched, and the new material will be
added right after them. So memcat probably begins with

void *target = (char*)s1 + n1;

... to get a pointer the the spot where the new material
will go. (The (char*) cast is needed because you can't do
arithmetic on a void* pointer.) What next? The rest of the
job is just copying the area 2 material to its new position.
So the complete implementation of memcat might look like

void *memcat(void *s1, size_t n1, void *s2, size_t n2) {
void *target = (char*)s1 + n1;
memcpy(target, s2, n2);
return s1;
}

... or, with some abbreviation
void *memcat(void *s1, size_t n1, void *s2, size_t n2) {
memcpy((char*)s1 + n1, s2, n2);
return s1;
}

(An "industrial-strength" version would probably use const
on s2, and a version for C99 compilers would also use the
restrict qualifier, but the outline would be the same.)

In other words, memcat is just memcpy with a different
starting point! And that's probably why it doesn't exist in
the Standard library: it's a trivial variation on a function
that's already provided. Adding it would be a little bit like
adding a sqrt_half function that computed the square root of
one-half its argument: A task that's easily done by calling
the usual sqrt function with a halved argument to begin with.

I tried your function as follows:-

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);

return s1;
}

int main(void)
{

int i =2,j=3,k;
int *p,*q,*r;

p = &i; q = &j;
r = mcat(p,sizeof(int),q,sizeof(int));

for(k=0;k<2;k++)
printf("\t %d",r[k]);
puts("");
return(EXIT_SUCCESS);
}

29,1
but getting the o/p as 2 1 instead of 2 3
 
M

Mark Bluemel

I tried your function as follows:-

Why? Eric pointed out that the function was stupid and unnecessary.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);

return s1;
}

int main(void)
{

int i =2,j=3,k;
int *p,*q,*r;

p = &i; q = &j;
r = mcat(p,sizeof(int),q,sizeof(int));

for(k=0;k<2;k++)
printf("\t %d",r[k]);
puts("");
return(EXIT_SUCCESS);
}

The crucial part of this code boils down to :-

int i =2,j=3,k;
void *source = &j;
void *target = ((char *)&i) + sizeof(int);
memcpy(target,source,sizeof(int));

Which could be expanded to

int i =2,j=3,k;
char *source = &j;
char *target = ((char *)&i) + sizeof(int);
for(k=0;k < sizeof(int);k++) {
*source++ = *target++;
}


Which in turn could be replaced with

int i =2,j=3,k;
int *target = &i + 1;
*target = j;

What makes you think you can assign to some arbitrary place in memory?
 
E

Eric Sosman

Mark said:
Why? Eric pointed out that the function was stupid and unnecessary.

No, I did not. I called it "a trivial variation on a
function that's already provided," but I never said it
was "stupid." If you have opinions, go ahead and state
them -- but state them as your own, not as mine.

Or as Shakespeare put it, "Who steals my parse steals
trash."
 
M

Mark Bluemel

Eric said:
No, I did not. I called it "a trivial variation on a
function that's already provided," but I never said it
was "stupid." If you have opinions, go ahead and state
them -- but state them as your own, not as mine.

Sorry about that - I was a little cavalier.
 
N

Nick Keighley

On 30 Nov, 03:42, (e-mail address removed) wrote:

<snip>

I know others have replied but I'm not sure if you understood
them
I tried your function as follows:-

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);

return s1;
}

int main(void)
{

int i =2,j=3,k;
int *p,*q,*r;

p = &i; q = &j;
r = mcat(p,sizeof(int),q,sizeof(int));

what is the point of p, q and r?

why not (ignoring for the moment that it doesn't work...)

mcat (&i, sizeof(int), &j, sizeof(int));

note r is uneccessary because j is copied to i + sizeof(int)

so why won't this work? Becuse you've go no space for the extra
bytes. Suppose an int is 4 bytes (I like concrete explanations!).
Thus i has been allocated 4 bytes. And so has has j. The mcat call
resolves to a memcpy call that tries to add 4 bytes onto the end
of a 4 byte object! In my country we talk of trying to put a quart
into a pint pot. (trying to put 1136ml of fluid into 568ml container).
If i were an array with two entries you might have more luck

int i[2] = {2, 0}, j = 3;
mcat (&i[0], sizeof(int), &j, sizeof(int));


--
Nick Keighley

Egon: Don't over run the end of the array
Venkman: Why?
Egon: It would be bad.
Venkman: I'm fuzzy on the whole good-bad thing. Whattya mean "bad?"
Egon: Try to imagine all life as you know it stopping instantaneously
and every molecule in your body exploding at the speed of light.
Ray: Total protonic reversal....
Venkman: Right, that's bad...OK.. important safety tip. Thanks, Egon.
(early Undefined Behaviour)
 
A

aarklon

On 30 Nov, 03:42, (e-mail address removed) wrote:

<snip>

I know others have replied but I'm not sure if you understood
them


I tried your function as follows:-

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);
return s1;
}
int main(void)
{
int i =2,j=3,k;
int *p,*q,*r;
p = &i; q = &j;
r = mcat(p,sizeof(int),q,sizeof(int));

what is the point of p, q and r?

why not (ignoring for the moment that it doesn't work...)

mcat (&i, sizeof(int), &j, sizeof(int));

note r is uneccessary because j is copied to i + sizeof(int)

so why won't this work? Becuse you've go no space for the extra
bytes. Suppose an int is 4 bytes (I like concrete explanations!).
Thus i has been allocated 4 bytes. And so has has j. The mcat call
resolves to a memcpy call that tries to add 4 bytes onto the end
of a 4 byte object! In my country we talk of trying to put a quart
into a pint pot. (trying to put 1136ml of fluid into 568ml container).
If i were an array with two entries you might have more luck

int i[2] = {2, 0}, j = 3;
mcat (&i[0], sizeof(int), &j, sizeof(int));

--

Okay i tried this:-

#include<string.h>
#include<stdio.h>
#include<stdlib.h>

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);

return s1;
}

int main(void)
{
int i[2] = {2, 0},j = 3;

printf("\t %d",i[0]);
printf("\t %d",i[1]);
printf("\t %d",i[2]);

mcat (&i[0], sizeof(int), &j, sizeof(int));

puts("");
printf("\t %d",i[0]);
printf("\t %d",i[1]);
printf("\t %d",i[2]);

puts("");
return(EXIT_SUCCESS);
}
but getting o/p as follows:-

2 0 -1208468512
2 3 -1208468512

instead of 2 0 3
 
S

santosh

On 30 Nov, 03:42, (e-mail address removed) wrote:

<snip>

I know others have replied but I'm not sure if you understood
them


I tried your function as follows:-

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);
return s1;
}
int main(void)
{
int i =2,j=3,k;
int *p,*q,*r;
p = &i; q = &j;
r = mcat(p,sizeof(int),q,sizeof(int));

what is the point of p, q and r?

why not (ignoring for the moment that it doesn't work...)

mcat (&i, sizeof(int), &j, sizeof(int));

note r is uneccessary because j is copied to i + sizeof(int)

so why won't this work? Becuse you've go no space for the extra
bytes. Suppose an int is 4 bytes (I like concrete explanations!).
Thus i has been allocated 4 bytes. And so has has j. The mcat call
resolves to a memcpy call that tries to add 4 bytes onto the end
of a 4 byte object! In my country we talk of trying to put a quart
into a pint pot. (trying to put 1136ml of fluid into 568ml
container). If i were an array with two entries you might have more
luck

int i[2] = {2, 0}, j = 3;
mcat (&i[0], sizeof(int), &j, sizeof(int));

--

Okay i tried this:-

#include<string.h>
#include<stdio.h>
#include<stdlib.h>

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);

return s1;
}

int main(void)
{
int i[2] = {2, 0},j = 3;

printf("\t %d",i[0]);
printf("\t %d",i[1]);
printf("\t %d",i[2]);

The valid range of indexes for the array 'i' are 0 and 1. You can point
to one past the end of an array, but you cannot deference it. Doing so
invokes undefined behaviour. So scrap that last printf() call.
mcat (&i[0], sizeof(int), &j, sizeof(int));

Wrong. The size of the array can be got by:

sizeof i

in this function. Passing it to another function causes it to "drop"
it's size information and sizeof will no longer work as above. It will
instead yield the size of the pointer to such an array. Therefore you
need to manually pass the size information.

Also 'i' needs to be at least one element larger to accomodate 'j' being
copied over.
puts("");
printf("\t %d",i[0]);
printf("\t %d",i[1]);
printf("\t %d",i[2]);

puts("");
return(EXIT_SUCCESS);
}
but getting o/p as follows:-

2 0 -1208468512
2 3 -1208468512

instead of 2 0 3

Make 'i' have three elements.
 
S

santosh

*Please* include context in your posts. I have added it in this
instance.
Okay i tried this:-

  #include<string.h>
  #include<stdio.h>
  #include<stdlib.h>

 void *mcat(void *s1, size_t n1, void *s2, size_t n2)
  {
    int k;
    void *target = (char*)s1 + n1;
    memcpy(target,s2, n2);

   return s1;
  }

  int main(void)
  {
    int i[2] = {2, 0},j = 3;

    printf("\t  %d",i[0]);
    printf("\t  %d",i[1]);
    printf("\t  %d",i[2]);
The valid range of indexes for the array 'i' are 0 and 1. You can
point to one past the end of an array, but you cannot deference it.
Doing so invokes undefined behaviour. So scrap that last printf()
call.
    mcat (&i[0], sizeof(int), &j, sizeof(int));
Wrong. The size of the array can be got by:

  sizeof i

in this function. Passing it to another function causes it to "drop"
it's size information and sizeof will no longer work as above. It
will instead yield the size of the pointer to such an array.
Therefore you need to manually pass the size information.
Also 'i' needs to be at least one element larger to accomodate 'j'
being copied over.
    puts("");
    printf("\t  %d",i[0]);
    printf("\t  %d",i[1]);
    printf("\t  %d",i[2]);
   
   puts("");
   return(EXIT_SUCCESS);
   }
   but getting o/p as follows:-
   
  2       0       -1208468512
  2       3       -1208468512
 
instead of 2 0 3
Make 'i' have three elements.

then whats the point in trying a mcat/memcat function ????.....

Point? That's something that you should know when you designed the
function. Mcat() as you have written it above expects the array at 's1'
to be "big enough" to attach the array starting at 's2' to it. If it is
not, then that constitutes a violation of contract by the caller and
results in a buffer overrun and consequent Undefined behaviour.

If you want the mcat() function to handle this situation then you need
to know the size of the array at 's1', not merely the offset at which
you are doing the concatenation. Then if the size is insufficient for a
straight copy, you need to dynamically enlarge the array using
realloc() before proceeding. This of course requires the original array
to be allocated with a previous call to *alloc(). Trying to resize a
static array is an error. This complicates the interface and usage of
the function significantly.

Actually your mcat() function as it stands can be completely replaced by
a single call to memcpy() or memmove(). So you might look into making
it handle undersized arguments to actually make it useful.
 
B

Barry Schwarz

On Fri, 30 Nov 2007 11:07:10 -0800 (PST), (e-mail address removed) wrote:


snip - It helps if you trim obsolete code from your posts.
Okay i tried this:-

#include<string.h>
#include<stdio.h>
#include<stdlib.h>

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);

return s1;
}

int main(void)
{
int i[2] = {2, 0},j = 3;

How many elements in the array i? What are their indices?
printf("\t %d",i[0]);
printf("\t %d",i[1]);
printf("\t %d",i[2]);

Do you really think i[2] exists? If you want to see a 3 in your
output, then print j.
mcat (&i[0], sizeof(int), &j, sizeof(int));

puts("");
printf("\t %d",i[0]);
printf("\t %d",i[1]);
printf("\t %d",i[2]);

puts("");
return(EXIT_SUCCESS);
}
but getting o/p as follows:-

2 0 -1208468512
2 3 -1208468512

instead of 2 0 3

Do you have any reason to believe that j follows i in memory?


Remove del for email
 
K

Kaz Kylheku

void *mcat(void *s1, size_t n1, void *s2, size_t n2)
{
int k;

Unused variable.
void *target = (char*)s1 + n1;
memcpy(target,s2, n2);

return s1;
}

This is just a flavor of memcpy where the destination is specified as
an address with displacement. In other words, it's just the memcpy
``instruction'' with a different ``addressing mode'' for the
destination operand! :)

How about indirect with scaled index and displacement, hahaha:

void *mcpy_ind_scidx_disp(void *base, size_t index, size_t scale,
size_t disp,
void *source, size_t size)
{
void *effective_addr = (char *) base + index * scale + disp;
return memcpy(effective_addr, source, size);
/* returning effective_addr is probably more useful than base or
source */
}

I hope you realize that the effective address calculation for the
addressing mode can be decoupled from the operation itself. That is to
say, we can separate the part which does the addressing mode into its
own function, and then just pass the resulting direct address to
regular memcpy.

void *ind_scidx_disp(void *base, size_t index, size_t scale)
{
return (char *) base + index * scale + disp;
}

Now instead of

mcpy_ind_scidx_disp(x, y, z, d, s);

you write:

memcpy(ind_scidx_disp(x, y, z), d, s);

and as a bonus, you can reuse the addressing mode calculation with
memset, memmove, and memchr.

It's exactly two characters extra to type, which is due to the extra
level of parentheses.
 

Ask a Question

Want to reply to this thread or ask your own question?

You'll need to choose a username for the site, which only take a couple of moments. After that, you can post your question and our members will help you out.

Ask a Question

Members online

Forum statistics

Threads
473,769
Messages
2,569,581
Members
45,056
Latest member
GlycogenSupporthealth

Latest Threads

Top