static_cast integer 地址到指针

Question

Why do you need a C-style cast for the following?

int* ptr = static_cast<int*>(0xff); // error: invalid static_cast from type 'int' 
                                    // to type 'int*'
int* ptr = (int*) 0xff; // ok.

Answer 1

static_cast can only cast between two related types. An integer is not related to a pointer and vice versa, so you need to use reinterpret_cast instead, which tells the compiler to reinterpret the bits of the integer as if they were a pointer (and vice versa):

int* ptr = reinterpret_cast<int*>(0xff);

Read the following for more details:

Type conversions

Answer 2

You need a C-style cast or directly the reinterpret_cast it stands for when casting an integer to a pointer, because the standard says so for unrelated types.

The standard mandates those casts there, because

1. you are doing something dangerous there.
2. you are doing something very seldom useful.
3. you are doing something highly implementation-dependent.
4. most times, that is simply a programming-error.

When should static_cast, dynamic_cast, const_cast and reinterpret_cast be used?
Regular cast vs. static_cast vs. dynamic_cast

Answer 3

Being late to the party I found that the following works only using static casts:

int* ptr1 = static_cast<int*>(static_cast<void*>(static_cast<unsigned char*>(nullptr) + 0xff));

This way you don't translate the constant directly to a pointer, instead you add it as a bytewise offset to the nullptr.