Printing addresses of variables

Question

When I run this code:

uint8_t stackVar = 0;
void* ptr = &stackVar;
uint8_t& ref = reinterpret_cast<uint8_t&>(ptr);
std::cout << (void*)&ref << std::endl;
std::cout << ptr << std::endl;
std::cout << (void*)&stackVar << std::endl;

I get this output:

0x22fe30
0x22fe3f
0x22fe3f

At least in my estimation I should get the same number for all three of these statements. What is going on here?

Answer 1

uint8_t& ref = reinterpret_cast<uint8_t&>(ptr);

You are casting a pointer ( void* ) to a reference. This will result in the same uint8_t , because it will make a reference to a temporary uint8_t , which you created out of a void pointer. 导致相同的uint8_t ，因为它将引用您从void指针创建的临时uint8_t 。 And because a new uint8_t is created, you are getting different addresses.

Maybe you meant uint8_t& ref = reinterpret_cast<uint8_t&>(stackVar);