What does this pointer notation mean?

Question

#define START ((void (**)(int)) 0x0fff)

*START = &fun_foo();

I haven't seen this before. What is happening here? Is void (**)(int) a function pointer?

score 3 · Accepted Answer · answered Oct 05 '12 at 05:00

3

void (**)(int) is a pointer to a pointer to a function that takes an int and returns nothing.

So START is apointer to a function pointer, and *START is the actual function pointer which is set to point to fun_foo.

answered Oct 05 '12 at 05:00

Andrew Cooper

score 1 · Answer 2 · edited May 23 '17 at 12:18

1

In your case, START is a pointer (located at the fixed address 0x0fff) to a function pointer.

But as I suggested in this answer, for readability reasons, you may want to use a typedef for the signature of that pointed function.

edited May 23 '17 at 12:18

Community

answered Oct 05 '12 at 04:59

2 Answers2