Storing an address into a function pointer

Question

I was watching this video about calling the bootloader using software by assigning the address of the beginning of the system memory into a function pointer and then calling it, the expression for storing the address :

sysMemBootJump = (void(*)(void))(*(u32*)0x1fff0004);

"sysMemBootJump" is the function pointer.

But what I don't understand is, why did he dereference the address of the memory before casting it to void(*)(void) ?

Answer 1

Because the 'real' entry point is stored at that address. Think of it as being a pointer-to-pointer-to-function, by dereferencing you get just the pointer-to-function.

Answer 2

This is equivalent to:

u32 ad = *(u32*)0x1fff0004;

^{this is fetching a word located at address 0x1fff0004}

then

sysMemBootJump = (void(*)(void))ad;

So 0x1fff0004 is the address of a word containing the routine's address.

And notice that the code is very unportable. A more portable integral type castable to/from addresses is uintptr_t from <stdint.h> ....

You can typedef function signatures, like here , to write more readable code.