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@xxxxxxxxxxxxxxxxxx>
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