Why member function address are so far away from free functions?

Question

Taking this example: https://godbolt.org/z/gHqCSA

#include<iostream>

template<typename Return, typename... Args>
std::ostream& operator <<(std::ostream& os, Return(*p)(Args...) ) {
    return os << (void*)p;
}

template <typename ClassType, typename Return, typename... Args>
std::ostream& operator <<(std::ostream& os, Return (ClassType::*p)(Args...) )
{
    unsigned char* internal_representation = reinterpret_cast<unsigned char*>(&p);
    os << "0x" << std::hex;

    for(int i = 0; i < sizeof p; i++) {
        os << (int)internal_representation[i];
    }

    return os;
}

struct test_debugger { void var() {} };
void fun_void_void(){};
void fun_void_double(double d){};
double fun_double_double(double d){return d;}

int main() {
    std::cout << "0. " << &test_debugger::var << std::endl;
    std::cout << "1. " << fun_void_void << std::endl;
    std::cout << "2. " << fun_void_double << std::endl;
    std::cout << "3. " << fun_double_double << std::endl;
}

// Prints:
//    0. 0x7018400100000000000
//    1. 0x100401080
//    2. 0x100401087
//    3. 0x100401093

I see the address of the member function is 0x7018400100000000000, which is understandable because member functions pointers have 16 bytes while free function as 0x100401080 have only 8 bytes.

However, why the member function address 0x7018400100000000000 is so far away from free function address 0x100401080? i.e., |0x7018400100000000000 - 0x100401080| = 0x70184000FFEFFBFEF80?

Why it is not closer i.e., something like 0x100401... instead of 0x701840...? Or I am printing the member function address wrong?

your mistake is thinking that member function addresses have anything to do with memory addresses — M.M, Jan 17 '20 at 01:19
printing a btye (`4`) as an `int` will be `"4"` instead of `"04"` needed to make sense of memory as hex, use `setw` and `setfill` as well — kmdreko, Jan 17 '20 at 01:26
A pointer-to-member-function is not a pointer. When you cast it to an `int` you get whatever is there, and having told the compiler to pretend that what's there is an `int`, you get what you asked for: a meaningless value. — Pete Becker, Jan 17 '20 at 14:37

score 3 · Accepted Answer · answered Jan 17 '20 at 01:42

3

Your architecture is little-endian. The low byte of the address is in the first byte of p, so your address is being printed out backwards.

answered Jan 17 '20 at 01:42

1201ProgramAlarm

32,384
7
42
56

And as @kmdreko pointed out in a comment on the question, without leading zeroes on any byte in the range `0x00` to `0x0f`. Reverse the output, and add a few zeroes in the obvious places, and the address of the member function is right next door to the others. – ShadowRanger Jan 17 '20 at 01:54
Maybe. But a pointer-to-member-function is not a pointer, so there's no sense in pretending that it is. – Pete Becker Jan 17 '20 at 14:40

Evandro Coan · Answer 2 · 2020-02-02T16:40:23.620

Fixed code which automatically detects little/big endian: https://godbolt.org/z/XSvT5R

#include <iostream>
#include <iomanip>
#include <sstream>

inline bool is_big_endian() {
    long int longvalue = 1;

    // https://stackoverflow.com/questions/8978935/detecting-endianness
    unsigned char* representation = reinterpret_cast<unsigned char*>(&longvalue);
    return ( (unsigned) representation[sizeof(long int) - 1] ) == 1;
}

template<typename Pointer>
std::ostream& print_pointer(std::ostream& os, const Pointer& pointer) {
    const unsigned char* representation = (unsigned char*) &pointer;

    int precision = 0;
    bool haszeros = false;

    unsigned firsthexdigit;
    unsigned secondhexdigit;

    std::ostringstream stream;
    stream.flags( os.flags() );
    stream << std::hex;

    #define print_pointer_HEX_DIGIT \
        firsthexdigit = (unsigned) representation[index] >> 4 & 0xf; \
        secondhexdigit = (unsigned) representation[index] & 0xf; \
        if( haszeros || firsthexdigit ) { \
            precision++; \
            haszeros = true ; \
            stream << firsthexdigit; \
        } \
        if( haszeros || secondhexdigit ) { \
            precision++; \
            haszeros = true ; \
            stream << secondhexdigit; \
        }

    if( is_big_endian() ) {
        for(int index = 0; index < static_cast<int>(sizeof pointer); index++) {
            print_pointer_HEX_DIGIT
        }
    }
    else {
        for(int index = static_cast<int>(sizeof pointer - 1); index >= 0 ; index--) {
            print_pointer_HEX_DIGIT
        }
    }

    if( os.precision() - ++precision > 0 ) {
        return os << "0x" + std::string( os.precision() - ++precision, '0' ) + stream.str();
    }

    return os << "0x" + stream.str();
}

template<typename Return, typename... Args>
std::ostream& operator <<(std::ostream& os, Return(*pointer)(Args...) ) {
    return print_pointer(os , pointer);
}

template <typename ClassType, typename Return, typename... Args>
std::ostream& operator <<(std::ostream& os, Return (ClassType::*pointer)(Args...) ) {
    return print_pointer(os , pointer);
}

struct test_debugger { void var() {} };
void fun_void_void(){};
void fun_void_double(double d){};
double fun_double_double(double d){return d;}

int main() {
    std::cout << "0. " << &test_debugger::var << std::endl;
    std::cout << "1. " << fun_void_void << std::endl;
    std::cout << "2. " << fun_void_double << std::endl;
    std::cout << "3. " << fun_double_double << std::endl;
    std::cout << "4. " << std::setfill('0') << std::setw(16) << fun_void_void << std::endl;
    std::cout << "5. " << std::setprecision(16) << fun_void_double << std::endl;
}

// Prints:
//    0. 0x100402e80
//    1. 0x100401118
//    2. 0x10040111f
//    3. 0x10040112b
//    4. 000000x100401118
//    5. 0x0000010040111f

Why member function address are so far away from free functions?

2 Answers2

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