CODEWITHSUNDEEP
functionpointersmemcpy
Abhishek Sagar
Abhishek Sagar

Reputation: 1326

copying address space of a function using memcpy

i have defined the fun pointer type as 

typedef void (*fn)(void)

declared a pointer variable : fn var = NULL; have written a small function void fun(void){} Now, i want to copy the address space of function fun into var variable using memcpy. I am trying this : memcpy(var, &fun, sizeof(fn))

But, when i printing the value of var it is showing up the garbage or negative value, and invoking the function from var is resulting into segmentation fault. pls help, where am i wrong here.

Upvotes: 0

Views: 1540

Answers (1)

Anton Kovalenko
Anton Kovalenko

Reputation: 21507

Your code is probably memcpy(&var,&fun,sizeof(fn)), or it would fail instantly with var initialized to NULL.

The memcpy above is copying first bytes of fun's code to var, which is not what you want (not that it's a defined behavior by any standard, it's just how it probably happens to misbehave on your OS/compiler).

Use just var = fun; instead.

If you're determined to play with memcpy (for fun or learning), you'll have to get a variable containing a pointer to function first, not just a value of that pointer:

fn var1 = fun;
fn var2 = NULL;
memcpy(&var2,&var1,sizeof(fn));
var2();

Upvotes: 1

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