Getting address of a member in a struct with dynamic libraries

Question

By looking at this answer: https://stackoverflow.com/a/4671482/1770034 I can use dlsym to get a global variable in C. Is it possible to get a member from a struct.

I'm guessing that if I had the following code in a shared object:

header.h

struct myStruct
{
    int a;
    int b;
};

imp.c

struct myStruct structure = { 123, 456 };

I could include the same header.h file and cast the whole pointer to a struct myStruct*. struct myStruct * capturedStructure = (struct myStruct*) dlsym(handle, "structure");

However, is there a way to get the address to member directly. I'm guessing I'm unable to do something like: int* c = (int*) dlsym(handle, "structure.b");

Since dlsym allows a person to grab a function or global by itself (without a header), I'm hoping that I can also grab a member without requiring a header.

Pretty sure the answer is "No." That's because the member offset calculations are handled in the code, and don't appear in the symbol table. And of course you can't rely on the offset being constant for all versions of the library. — user3386109
– user3386109, Commented Jan 3, 2019 at 0:08
@user3386109 - Does that mean that for a given arch (ie: x86, x86_64) I am unable to use int* b = (int*) (dlsym(handle, "structure") + 4); I suppose the answer is "No" because the compiler may reorder my members to reduce padding between members. — Questionable
– Questionable, Commented Jan 3, 2019 at 0:13
Using fixed offsets won't work in the long run. For example version 1 of the library has int a; int b; Then version 2 of the library adds another member int a; int c; int b;. The offset to b will change. So even if you have the header for the structure, you need to use the correct version of the header for each version of the library. — user3386109
– user3386109, Commented Jan 3, 2019 at 0:16
BTW, although the compiler may put padding between members, it is not allowed to reorder members. — user3386109
– user3386109, Commented Jan 3, 2019 at 0:18
@user3386109 - Thank you I didn't realize that. I thought that you needed to use __attribute__((packed)) [in gcc] to prevent reordering, but that attribute really only prevents padding from being inserted and you assume responsibility of aligning the data correctly. — Questionable
– Questionable, Commented Jan 3, 2019 at 0:24

KamilCuk · Accepted Answer · 2019-01-03 00:25:22Z

2

the address to member directly

The normal way looks like this:

struct myStruct *pnt = (struct myStruct*) dlsym(handle, "structure");
int *b = &pnt->b;

Now let's substitute s/pnt/((struct myStruct*) dlsym(handle, "structure"))/. That's:

int *b = &((struct myStruct*) dlsym(handle, "structure"))->b;

without the compiler having the structure defined? (from comments)

This could be a bit tricky, but we can do it. You would need to export another symbol:

const size_t offsetof_member_b_in_myStruct = offset(struct myStruct, b);

And then later in client code:

int *b = (int*)(
             (uintptr_t)dlsym(handle, "structure") + 
             *(size_t*)dlsym(handle, "offsetof_member_b_in_myStruct")
         );

I guess such API could be consistent, but feels bad. It's simpler to export the structure to client code. Maybe in a general case, better create a standard specifying memory layout of the structure you are interchanging with client code (thus you push the responsibility to clients to provide proper abstraction).

edited Jan 3, 2019 at 0:25

answered Jan 2, 2019 at 23:56

KamilCuk

146k8 gold badges84 silver badges154 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

Questionable Over a year ago

This will work, but does the symbol table keep track of the offsets so that I can access the member "b" without the compiler having the structure defined? I can manually do int* b = (int*) (dlsym(handle, "structure") + 4);

KamilCuk Over a year ago

without the compiler having the structure defined? - you can, but notice that this is wrong. Read about padding in structures. You can export offsetof(struct myStruct, b) in a variable and then (int*)((uintptr_t)dlsym(handle, "structure") + offsetof(struct myStruct, b)) but it feels crude. It is wrong to assume that offsetof(struct myStruct, b) is equal to 4.

Questionable Over a year ago

Thank you. Yes, I didn't consider that the compiler may reorder the members to reduce padding between the elements.

Ctx Over a year ago

@Questionable A C compiler may not reorder members of a struct.

user3386109 Over a year ago

To me it seems a lot cleaner to export a "getter" that returns the value of b.

|

Petr Skocik · Accepted Answer · 2019-01-02 23:57:01Z

1

With extern struct myStruct structure; there will be an entry in the symbol table pointing to structure and it will be keyed to the string structure.

To get the address of its b member, you simply do:

struct myStruct *p = dlsym(handle, "structure");
if(!p) fail();
int *pb = &p->b;

answered Jan 2, 2019 at 23:57

Petr Skocik

60.6k7 gold badges105 silver badges165 bronze badges

Collectives™ on Stack Overflow

Getting address of a member in a struct with dynamic libraries

2 Answers 2

7 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

7 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related