Assign function pointer to a function using the address of function pointer variable

Question

I have a function pointer defined like below:

typedef void (*FPT)(void);
FPT Fp;

The pointer variable "Fp" is located at address 0x1234 I have my function defined like below:

void myfunc (void)
{
    return;
}

I do not have access to the symbol name "Fp" but I know its address (0x1234). Now how do I assign the address of myfunc() to "Fp" ??

kaylum · Accepted Answer · 2015-04-16 23:16:03Z

1

This should do what you want.

FPT *fpt_pointer = (FPT *)0x1234;

*fpt_pointer = myfunc;

But note that it's very bad practice to have hard coded addresses like this. I don't know whether you really intend to do that (don't!) or whether you are just describing it that way to simplify the question. It is very likely to break the next time you run the program (even without a re-compile due to Address Space Layout Randomisation).

answered Apr 16, 2015 at 23:16

kaylum

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3 Comments

Jonathan Leffler Over a year ago

Not to mention 0x1234 being a fairly implausible address for a function.

user3125367 Over a year ago

@JonathanLeffler It was bogus, ofc. *(FPT *)offset = myfunc; is fine too.

Santosh Over a year ago

This is exactly what I am trying, just wanted to see if there are other ways. And yes, it is just a random address I typed in. I definitely dont intent to put a hard coded address. It is just easier to explain the question here. Thanks guys

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