memory address represented as a string

memory address represented as a string

[Saurabh Sehgal]

Hi all,

I had a quick question:

Let's say I design a function with the signature:

void * foo( char * addr ) ; ,

where addr is a string that represents a valid memory address  ...
so the way someone can call this function is ...

char * addr = "0xae456778" // assume this is a valid memory address on
the machine
foo( addr ) ;

Is it possible to take this address in string form, and assign it to
an actual pointer of void * type ?
I want the function "foo" to return a pointer pointing to the memory
location as indicated
by the string passed in.

Thank you !

Saurabh

Re: memory address represented as a string

[Aneesh Bhasin]

Hi Saurabh,

On Sun, Jul 26, 2009 at 1:09 PM, Saurabh Sehgal <saurabh.r.s@gmail.com> wrote:
>
> Hi all,
>
> I had a quick question:
>
> Let's say I design a function with the signature:
>
> void * foo( char * addr ) ; ,
>
> where addr is a string that represents a valid memory address  ...
> so the way someone can call this function is ...
>
> char * addr = "0xae456778" // assume this is a valid memory address on
> the machine

This means that addr points to a memory location where the string
stored is "0xae456778"


>
> foo( addr ) ;
>
> Is it possible to take this address in string form, and assign it to
> an actual pointer of void * type ?

Yes, you can parse each individual character of the hex-format string
and convert it to its integral equivalent (you can find many samples
of this on internet...) and assing this value to a void * and then
return it back..

Hope that helps..
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Re: memory address represented as a string

[Manish Katiyar]

On Sun, Jul 26, 2009 at 1:29 PM, Aneesh Bhasin<contact.aneesh@gmail.com> wrote:
> Hi Saurabh,
>
> On Sun, Jul 26, 2009 at 1:09 PM, Saurabh Sehgal <saurabh.r.s@gmail.com> wrote:
>>
>> Hi all,
>>
>> I had a quick question:
>>
>> Let's say I design a function with the signature:
>>
>> void * foo( char * addr ) ; ,
>>
>> where addr is a string that represents a valid memory address  ...
>> so the way someone can call this function is ...
>>
>> char * addr = "0xae456778" // assume this is a valid memory address on
>> the machine
>
> This means that addr points to a memory location where the string
> stored is "0xae456778"
>
>
>>
>> foo( addr ) ;
>>
>> Is it possible to take this address in string form, and assign it to
>> an actual pointer of void * type ?

/tmp> cat test.c
#include<stdio.h>

int *addr(char *address)
{
  int ptr;
  ptr=strtoul(address, 0 , 16);
  if (ptr==-1)
    {
      perror("strtoul failed:");
    }
  return (int *)ptr;
}

int *foo()
{
  char c[10];
  int *ptr=malloc(sizeof(int));
  sprintf(c,"%p",ptr);
  *ptr=100;
  return addr(c);

}

int main()
{
  int *p=foo();
  printf("%d\n",*p);
  return 0;
}
/tmp> gcc test.c
test.c: In function ‘foo’:
test.c:17: warning: incompatible implicit declaration of built-in
function ‘malloc’
/tmp> ./a.out
100

Thanks -
Manish


>
> Yes, you can parse each individual character of the hex-format string
> and convert it to its integral equivalent (you can find many samples
> of this on internet...) and assing this value to a void * and then
> return it back..
>
> Hope that helps..
> --
> To unsubscribe from this list: send the line "unsubscribe linux-c-programming" in
> the body of a message to majordomo@vger.kernel.org
> More majordomo info at  http://vger.kernel.org/majordomo-info.html
>



-- 
Thanks -
Manish
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Re: memory address represented as a string

[Saurabh Sehgal]

Hi all,

Thanks for the reply !

I was able to write this function successfully.

This would lead me to my next question.

Is there anyway to test if a memory address is valid or not without causing a
segmentation fault and catching this maybe in a signal handler ?

Is there a safe way in the same function " void * foo (char * addr)"
to check that
if the address contained in the string represented by "char * addr"  is valid
before returning it.

Thank you !

Saurabh

On Sun, Jul 26, 2009 at 4:14 AM, Eric Polino<aluink@gmail.com> wrote:
> sure, you can just write a function that declares a void * and uses it as a
> regular numeric type such as int or whatnot and parses the string
> representation as a numeric value into that variable.  You can do stuff the
> same as you would with any other integer type, "x += 10; x -= 18;".  So the
> function would look something like
>
> void *foo(const char *addr){
>   void *parsed_value = 0;
>   ....
>   /* Parse addr into parsed_value */
>   ....
>   /* parsed_value == value_described_by(addr) */
>
>   return parsed_value;
> }
>
> Parsing addr into parsed_value is left as an exercise to the reader ;)
>
> "None are more hopelessly enslaved than those who falsely believe they are
> free."
>                                      --Goethe
>
> "Freedom is living without government coercion."
>                   --Ron Paul (www.ronpaul2008.com)
>
>
> On Sun, Jul 26, 2009 at 03:39, Saurabh Sehgal <saurabh.r.s@gmail.com> wrote:
>>
>> Hi all,
>>
>> I had a quick question:
>>
>> Let's say I design a function with the signature:
>>
>> void * foo( char * addr ) ; ,
>>
>> where addr is a string that represents a valid memory address  ...
>> so the way someone can call this function is ...
>>
>> char * addr = "0xae456778" // assume this is a valid memory address on
>> the machine
>> foo( addr ) ;
>>
>> Is it possible to take this address in string form, and assign it to
>> an actual pointer of void * type ?
>> I want the function "foo" to return a pointer pointing to the memory
>> location as indicated
>> by the string passed in.
>>
>> Thank you !
>>
>> Saurabh
>> --
>> To unsubscribe from this list: send the line "unsubscribe
>> linux-c-programming" in
>> the body of a message to majordomo@vger.kernel.org
>> More majordomo info at  http://vger.kernel.org/majordomo-info.html
>
>



-- 
Saurabh Sehgal
E-mail:     saurabh.r.s@gmail.com
Phone:     647-831-5621
LinkedIn: http://www.linkedin.com/pub/1/7a3/436
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Re: memory address represented as a string

[Rahul K Patel]

Hi,

Saurabh Sehgal wrote:
> Hi all,
>
> Thanks for the reply !
>
> I was able to write this function successfully.
>
> This would lead me to my next question.
>
> Is there anyway to test if a memory address is valid or not without causing a
> segmentation fault and catching this maybe in a signal handler ?
>
> Is there a safe way in the same function " void * foo (char * addr)"
> to check that
> if the address contained in the string represented by "char * addr"  is valid
> before returning it.
>   

static inline int is_invalid_addr( const void *p )
{
       int dev_null_fd = open("/dev/null", O_RDWR);
       if(dev_null_fd < 0)
          return -1;
   
        /* HACK: on some systems we can not write to check for pointer 
validity */
        size_t ret = fcntl( dev_null_fd, F_GETLK, p );

        ret = ( errno == EFAULT );
        errno = 0;
        return ret;
}

You can use this function to check whether given address is accessible.

-- Rahul Patel.
> Thank you !
>
> Saurabh
>
> On Sun, Jul 26, 2009 at 4:14 AM, Eric Polino<aluink@gmail.com> wrote:
>   
>> sure, you can just write a function that declares a void * and uses it as a
>> regular numeric type such as int or whatnot and parses the string
>> representation as a numeric value into that variable.  You can do stuff the
>> same as you would with any other integer type, "x += 10; x -= 18;".  So the
>> function would look something like
>>
>> void *foo(const char *addr){
>>   void *parsed_value = 0;
>>   ....
>>   /* Parse addr into parsed_value */
>>   ....
>>   /* parsed_value == value_described_by(addr) */
>>
>>   return parsed_value;
>> }
>>
>> Parsing addr into parsed_value is left as an exercise to the reader ;)
>>
>> "None are more hopelessly enslaved than those who falsely believe they are
>> free."
>>                                      --Goethe
>>
>> "Freedom is living without government coercion."
>>                   --Ron Paul (www.ronpaul2008.com)
>>
>>
>> On Sun, Jul 26, 2009 at 03:39, Saurabh Sehgal <saurabh.r.s@gmail.com> wrote:
>>     
>>> Hi all,
>>>
>>> I had a quick question:
>>>
>>> Let's say I design a function with the signature:
>>>
>>> void * foo( char * addr ) ; ,
>>>
>>> where addr is a string that represents a valid memory address  ...
>>> so the way someone can call this function is ...
>>>
>>> char * addr = "0xae456778" // assume this is a valid memory address on
>>> the machine
>>> foo( addr ) ;
>>>
>>> Is it possible to take this address in string form, and assign it to
>>> an actual pointer of void * type ?
>>> I want the function "foo" to return a pointer pointing to the memory
>>> location as indicated
>>> by the string passed in.
>>>
>>> Thank you !
>>>
>>> Saurabh
>>> --
>>> To unsubscribe from this list: send the line "unsubscribe
>>> linux-c-programming" in
>>> the body of a message to majordomo@vger.kernel.org
>>> More majordomo info at  http://vger.kernel.org/majordomo-info.html
>>>       
>>     
>
>
>
>   
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Re: memory address represented as a string

[Glynn Clements]

Saurabh Sehgal wrote:

> I had a quick question:
> 
> Let's say I design a function with the signature:
> 
> void * foo( char * addr ) ; ,
> 
> where addr is a string that represents a valid memory address  ...
> so the way someone can call this function is ...
> 
> char * addr = "0xae456778" // assume this is a valid memory address on
> the machine
> foo( addr ) ;
> 
> Is it possible to take this address in string form, and assign it to
> an actual pointer of void * type ?
> I want the function "foo" to return a pointer pointing to the memory
> location as indicated
> by the string passed in.

Yes:

	const char *addr = "0xae456778";
	void *ptr;

	sscanf(addr, "%p", &ptr);

The string needs to be in the format used by printf("%p"), which is
platform-specific.

Alternatively:

	ptr = (void *) strtoull(addr, NULL, 16);

However, strtoull() isn't in C89, although it's in C99 and POSIX.

Using strtoul() will work if sizeof(long) >= sizeof(void *). AFAIK,
this is true on all versions of Linux, but may not be true on some
other 64-bit platforms. On particular, it isn't true on Win64, where
"long" is only 32 bits.

-- 
Glynn Clements <glynn@gclements.plus.com>