Convert string formatted memory address to pointer in C

Question

I have a text file with the memory addresses of certain components of a program. For instance part of that text file looks like this:

00400000-0040b000

or

7f9ae612f000-7f9ae62ee000

I have already read the file into my program and assigned a char[] with the first address (00400000).

I also have a structure with a null pointer

struct myStruct
{
    void *address;
}

And I would like the pointer *address to have the value of 0x00400000, how would I do this?

Thank you for your help.

It's (apparently) simpler than you think: `address = (void *) 0x00400000` — Some programmer dude, Jan 08 '17 at 02:06
@Someprogrammerdude He wants to read it `string-formatted`. You can't just cast that to `void *` — , Jan 08 '17 at 12:00

Remy Lebeau · Accepted Answer · 2017-01-08T07:52:38.890

1

Use sscanf() to parse the string as a hex integer and then typecast the resulting integer to a void* pointer. For example:

char str[] = "00400000";
unsigned long ul;
struct myStruct s;

sscanf(str, "%lx", &ul);
s.address = (void*) (uintptr_t) ul;

edited Jan 08 '17 at 07:52

answered Jan 08 '17 at 02:07

Remy Lebeau

555,201
31
458
770

Wouldnt %p do this directly? – Vality Jan 08 '17 at 02:19
I now get an error listed below "cast to pointer from integer of different size [-Wint-to-pointer-cast] memReg.startAddr = (void*) i;" – BionicSheep Jan 08 '17 at 02:26
@BionicSheep: Your addresses look like 32-bit values, but your compiler seems to be complaining about different sizes. What size are your `unsigned int` and `void *`? – Jonathan Leffler Jan 08 '17 at 02:33
2

use %lu and uintptr_t – Jan 08 '17 at 02:53
@MarkYisri that works with the errors but for any addresses with letters in them it fails – BionicSheep Jan 08 '17 at 02:59
2

sorry, %lx @bionicsheep – Jan 08 '17 at 03:06
@Vality although `%p` is indeed meant for pointers, its format is platform/implementation specific, so it is not suitable for the OP's example. – Remy Lebeau Jan 08 '17 at 03:09
ill post as an answer later. – Jan 08 '17 at 03:14
`"%lx"` is not the specified matching specifier for `uintptr_t`. – chux - Reinstate Monica Jan 08 '17 at 04:11

score 0 · Answer 2 · answered Jan 08 '17 at 02:40

If the same instance of the program printed "00400000-0040b000" then reading a void* back via scanf("%p"... is possible. Else the result is undefined.

p specifier:

Matches an implementation-deﬁned set of sequences, which should be the same as the set of sequences that may be produced by the %p conversion of the fprintf function. The corresponding argument shall be a pointer to a pointer to void. The input item is converted to a pointer value in an implementation-deﬁned manner. If the input item is a value converted earlier during the same program execution, the pointer that results shall compare equal to that value; otherwise the behavior of the %p conversion is undeﬁned. C11dr §7.21.6.2 12

struct myStruct x;
if (fscanf(stream, "%p", &x.address) == 1) Success();

score 0 · Answer 3 · edited Jun 20 '20 at 09:12

Source:

First of all %x writes unsigned int. There is no guarantee that pointer size = unsigned int size.

As there is no format specifier which takes uintptr_t, you need to manualy check your pointer size in program and then select correct format specifier.

Therefore, to comply with the standard, you have to manually use the sizeof operator on void * and unsigned int and compare them. If they are equal, you can use %x as the format.

Otherwise, you need to take appropriate action, and use a different format.

Of course, if this is just a test project, it won't matter to do a dynamic check at runtime. Just hardcode it to the correct value. Be sure to leave a FIXME though, so you can come back and add dynamic checking later.

Additionally,

We discussed in the comments about the format specifiers, and you specified that %lx had correct results. Since it is guarenteed that %lx can print a value of at least uint32_t, your void * must be at least the size of uint32_t.

Additionally number 2,

Source.

If you aren't happy with the implementation-defined behaviour of %p, then use C99 <inttypes.h> and uintptr_t instead:

printf("0x%" PRIXPTR "\n", (uintptr_t)your_pointer);

Please note that the above solution requires C99 to work properly.

Convert string formatted memory address to pointer in C

3 Answers3