What is (DWORD*) in C++

In C++, the notation (DWORD*) is a pointer to a data type called DWORD. DWORD stands for Double Word and is used to store a value that's an unsigned 32-bit integer. The * operator before a data type creates a pointer to that data type. So in this case, (DWORD*) creates a pointer to a DWORD. Pointers are commonly used in C++ to keep track of memory addresses, and they can be used to directly manipulate memory, like when working with dynamic memory allocation or other advanced tasks.

When you output a pointer it gives you sizeof(void*) bytes. If you're on x64 this is 16 digits 

@_realnickk

No. Simply not, study where "WORD" came from and you'll understand why DWORD is not 8 bytes on x64 lol, WORD is historically known for being 16 bits, DWORD is always 32 bits on both x86 and AMD64/x86-64. And to answer the guys, question, when you std::cout << ptr, a operator<< overload which takes in a std::ostream& and a const void* will be called which will output a hexadecimal value, for example:

std::cout << reinterpret_cast< DWORD* >( 313231 ) will result in the following output: 000000000004C78F.

There aren't "more zeros", the overload just tries to be accurate with the output instead of outputting it as a regular integer.

Also, Nick, your explanation was cloudy at best, I'm not going to bother explaining pointers here as I have already made a tutorial on them but, when you explain concepts your teminology should be at least coherent.

@_realnickk

Typically it doesn't, most implementations don't change it, though it is standardized that an unsigned long must be at minimum 4 bytes/32 bits, though any implementation can go above that, most don't though. This also goes for other integer data types where short is minimum 16 bits, long is minimum 32 bits, long long is minimum 64 bits.

@_realnickk

Yes