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How to transfer an address value without using pointers

Hi,

I have the following program structure:

main
.....
int A=5;
int* ptr= A; (so at this point ptr stores address of A)
.....
.....
send_data(...)
....
.....

for the function send_data, I need on its body to use the address of
A:
void send_data(...){
.....
.....
char* R=(char*) ptr;
......

but due to the limitations of the compiler (please dont ask me why :))
I can not use pointers here.

At the moment I do not know how to send to the function the address
information, somebody has idea ?

Thanks in advance
Jun 27 '08 #1
69 3347
On Jun 16, 10:39 pm, Horacius ReX <horacius....@g mail.comwrote:
Hi,

I have the following program structure:

main
....
int A=5;
int* ptr= A; (so at this point ptr stores address of A)
Don't you mean &A?
....
....
send_data(...)
send_data?
...
....

for the function send_data, I need on its body to use the address of
A:
void send_data(...){
....
....
char* R=(char*) ptr;
.....

but due to the limitations of the compiler (please dont ask me why :))
I can not use pointers here.
Why?
At the moment I do not know how to send to the function the address
information, somebody has idea ?
You can define ptr to belong to the global scope.
Jun 27 '08 #2
On 16 Jun, 20:39, Horacius ReX <horacius....@g mail.comwrote:
but due to the limitations of the compiler (please dont ask me why :))
I can not use pointers here.
You may well get better answers if you explain why you can't use
pointers. Is it a problem with the compiler or with your understanding
of the compiler?
Jun 27 '08 #3
Horacius ReX wrote:
Hi,

I have the following program structure:

main
....
int A=5;
int* ptr= A; (so at this point ptr stores address of A)
No. You want:

int *ptr = &A;
....
....
send_data(...)
...
....

for the function send_data, I need on its body to use the address of
A:
void send_data(...){
....
....
char* R=(char*) ptr;
.....
This assigns to R the value derived from interpreting the contents of
ptr as a char * value.
but due to the limitations of the compiler (please dont ask me why :))
I can not use pointers here.

At the moment I do not know how to send to the function the address
information, somebody has idea ?

Thanks in advance
If your system is this limited then you might try storing the address in
a appropriate integer object. If your system has them use intptr_t or
uintptr_t.

Jun 27 '08 #4
Horacius ReX wrote:
Hi,

I have the following program structure:

main
.....
int A=5;
int* ptr= A; (so at this point ptr stores address of A)
No, ptr contains the value of A, which is 5.
To store the address of A you need
int *ptr = &A;

I suspect your subsequent problems are related to this basic error.
Jun 27 '08 #5
yes, i tried tranferring an int with the address "coded", but dont
know how to use later in the function this address
Something like this?

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

void send(int ptr){
*(int*)ptr = 1201;
}

int main(void) {
int A=0;

send((int)&A);

printf("A=%d\n" ,A);
}
what do you mean with intptr_t and uintptr_t ?
Special int types guaranteed to be the right size for your pointers.
If you need your code to run on every computer ever made, past,
present and future, then you need to worry about stuff like this.

-- Bartc
Jun 27 '08 #6
On Jun 17, 12:15 am, Bart <b...@freeuk.co mwrote:
yes, i tried tranferring an int with the address "coded", but dont
know how to use later in the function this address

Something like this?

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

void send(int ptr){
*(int*)ptr = 1201;

}

int main(void) {
int A=0;

send((int)&A);

printf("A=%d\n" ,A);

}

thanks! this is really what was looking for !
Jun 27 '08 #7
Martin Ambuhl <ma*****@earthl ink.netwrites:
Horacius ReX wrote:
>Hi,
I have the following program structure:
main
.....
int A=5;
int* ptr= A; (so at this point ptr stores address of A)

No, ptr contains the value of A, which is 5.
No, the declaration violates a constraint, so ptr could contain
anything at all assuming the translation unit isn't rejected.

[...]

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Jun 27 '08 #8
Horacius ReX <ho**********@g mail.comwrites:
On Jun 16, 10:03 pm, gw7...@aol.com wrote:
>On 16 Jun, 20:39, Horacius ReX <horacius....@g mail.comwrote:
but due to the limitations of the compiler (please dont ask me why :))
I can not use pointers here.

You may well get better answers if you explain why you can't use
pointers. Is it a problem with the compiler or with your understanding
of the compiler?

it is a grade question, :)
Do you mean that this is a homework question?

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Jun 27 '08 #9
On Jun 16, 11:15*pm, Bart <b...@freeuk.co mwrote:
Something like this?

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

void send(int ptr){
* *(int*)ptr = 1201;

}
If you compile this for MacOS X 64 bit (and probably some other
operating systems), this is not just likely to crash, it is one
hundred percent guaranteed to crash.
Jun 27 '08 #10

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