你如何获得函数的原始地址？

Question

I'm writing a Jit compiler and I need to call a function based on its (offset) address from assembly. IE, given:

void myfunction() { }

I want to be able to get the actual address for this function so I can call it by offset from assembly. I've tried a couple of things, but I get different answers:

void *address1 = (void *&)myfunction;
void (*address2)() = &myfunction;

cout << &address1 << endl; // 0x7fff5fbff800
cout << &(*address1) << endl; // 0x20ec8348e5894855
cout << &(address2) << endl; // 0x7fff5fbff808

Which of those is correct (if either of them is)? If there is an easier way I'm overlooking, definitely interested in that as well.

Update

I did the smart thing and did an objdump , which says:

000000010000fd70 <__Z10myfunctionv>:
10000fe76:  48 8d 05 f3 fe ff ff    lea    -0x10d(%rip),%rax        # 10000fd70 <__Z10myfunctionv>

so the correct answer appears to be:

cout << (void *)(void (*)())myfunction << endl; // 0x10000fd70

Per Eric below, this is the best answer:

cout << hex << (uintptr_t)myfunction << endl;

Answer 1

There is no standard way to do this in C++, because the standard does not define any conversions from pointers to integers or to non-function pointers or any behavior for printing function pointers other than interpreting them as true/false, and function pointers are always true (non-null). Function pointers are not intended to be printed or exported from a program.

In C, you may convert a function pointer to uintptr_t and print that:

#include <inttypes.h>
#include <stdint.h>
#include <stdio.h>
…
    printf("0x%" PRIxPTR "\n", (uintptr_t) myfunction);

Some C++ implementations will allow you to cast function pointers to other types of pointers or to integer types. You should ensure the conversions performed by these casts are supported by the C++ implementation(s) you are using before relying on them. In some implementations, the value of a function pointer might not be the address of actual code; it might be the address of a descriptor of the function. In that case, branching to the address would not work; you need to conform to the platform's requirements for calling functions.

Printing the address of a function for further use in other programs is rarely likely to be of any use. Using the printed address to call the function typically requires recompiling the program with new code that uses the address derived from the printing, and changing the program and then recompiling may change the address of the function. The usual way that a function would be called from assembly is that its address would be provided to assembly code other via the normal linking mechanisms (refer to it by name in assembly and let the linker resolve the reference) or by passing a pointer to the function to the assembly routine as a parameter.

Another option is to place the function in a dynamic library, in which case its address can be looked up by features of the dynamic loader, such as the dlsym routine.

Answer 2

To get the memory address of a function, use the function name without parentheses.

void f(); 
void (*p_fun)() = f;

You may optionally include an ampersand:

void f(); 
void (*p_fun)() = & f;

So, for your question, address2 is pointing to the function, to print out its address, use

cout << address2 << endl;

Answer 3

The correct one is

 cout << &(*address1) << endl; // 0x20ec8348e5894855

the same would be

 cout << address1 << endl; 

The other two are referring to the addresses of the function pointer, ie the variable adress1 and address2.