CODEWITHSUNDEEP
c++gccfunction-pointers
code shogan
code shogan

Reputation: 839

why is the address of a c++ function always True?

well why would,

#include <iostream>

using namespace std;

int afunction () {return 0;};

int anotherfunction () {return 0;};

int main ()
{
    cout << &afunction << endl;
}

give this,

1

  1. why is every functions address true?
  2. and how then can a function pointer work if all functions share (so it seems) the same addresss?

Upvotes: 6

Views: 1185

Answers (7)

Dennis Zickefoose
Dennis Zickefoose

Reputation: 10969

The function address isn't "true". There is no overload for an ostream that accepts an arbitrary function pointer. But there is one for a boolean, and function pointers are implicitly convertable to bool. So the compiler converts afunction from whatever its value actually is to true or false. Since you can't have a function at address 0, the value printed is always true, which cout displays as 1.

This illustrates why implicit conversions are usually frowned upon. If the conversion to bool were explicit, you would have had a compile error instead of silently doing the wrong thing.

Upvotes: 15

Robᵩ
Robᵩ

Reputation: 168626

There is no overloaded function: operator<<(ostream&, int(*)()), so your function pointer is converted into the only type that works, bool. Then operator<<(ostream&, bool) is printing the converted value: 1.

You may be able to print the function address like so:

cout << (void*)&afunction << endl;

Upvotes: 1

user2100815
user2100815

Reputation:

There cannot be an overload for function pointers for the iostream << operator, as there are an infinite number of possible function pointer types. So the function pointer gets a conversion applied, in this case to bool. Try:

cout << (void *) afunction << endl;

Which will give you the address in hex - for me the result was:

0x401344

Upvotes: 0

Steve Jessop
Steve Jessop

Reputation: 279255

There's no overload of operator<< for function pointers (except stream manipulators), but there is one for bool, so the function pointer is converted to that type before display.

The addresses aren't equal, but they're both non-null, and hence they both covert to true.

Upvotes: 2

Nikolai Fetissov
Nikolai Fetissov

Reputation: 84159

The function pointer type is not supported by std::ostream out of the box. Your pointers are converted to only possible compatible type - bool - and verything that is not zero is true thanks to backward compatibility to C.

Upvotes: 3

mystery
mystery

Reputation: 19513

Did you check anotherfunction() as well?

Anyway, C++ pointer addresses, like C pointer addresses, are usually virtual on most platforms and don't correspond directly to memory locations. Hence the value may be very small or unusual.

Also, they will always be true, as 0 is NULL, an invalid pointer, and anything that is over 0 is true

Upvotes: -1

Mark Ransom
Mark Ransom

Reputation: 308138

All addresses in C++ are non-zero, because zero is the NULL pointer and is a reserved value. Any non-zero value is considered true.

Upvotes: 0

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