如果函数指针与数组地址一起传递会发生什么

Question

typedef void(*FUNC)(void);
int main(void)
{
        //intptr_t m;
    const static unsigned char insn[4] = { 0xff, 0xff, 0xff, 0xff };
        FUNC function = (FUNC) insn;
        function();
}

The above code gives me an output as Illegal instruction. Can someone explain why ? . Is it because the function pointer is not having the address of a function (as its having the address of an array it couldnot jump to the address)

Answer 1

Since the pointer to the first element of the array is not a pointer to a function, you invoke undefined behaviour by calling a 'function' via the variable function . When you invoke undefined behaviour, anything can happen. A crash with an illegal instruction is perfectly legitimate; so is wiping all the data off your disk.

There is nothing that can be 'expected' according to the standards. As hinted in the comments, what is likely to happen is that the bytes stored on the stack in the array insn (and in the rest of the stack, with the stack frame for main() and things like the argument list and the environment variables) will be treated as machine code. Fortunately for you, one of the bytes is an invalid (or illegal) instruction, and the program stops.

Answer 2

C11 J.5.7 Function pointer casts

1 A pointer to an object or to void may be cast to a pointer to a function, allowing data to be invoked as a function (6.5.4).

2 A pointer to a function may be cast to a pointer to an object or to void , allowing a function to be inspected or modified (for example, by a debugger) (6.5.4).

You are fine if you only cast the array name to a function pointer, but it's undefined behavior that you call it because it's not actually a function.

You can try storing actual function pointer addresses in the array and then cast it back to function pointer and call them. However, unsigned char still won't work because it's too small, use uintptr_t instead.