如何将int转换为LPBYTE

Question

I need to convert int value to LPBYTE. when I look at the definitions is shows like this. I'm not sure what is far word says.

typedef unsigned char       BYTE;
typedef BYTE far            *LPBYTE;

1. What is the meaning of 'far'
2. How to convert int value to LPBYTE

Edit

foo(LPBYTE x){
}

int main()
{
 int y = koo();
 foo(y); // how to cast here
 return 0;
}

Actual code

 int iVal = 0;
 LONG res = RegQueryValueEx(hKey, L"UseSystemSeparators", NULL, &lpType, (LPBYTE)iVal, &size);

Answer 1

If I understand you correctly, you want the address of the variable , and then convert that address to LPBYTE .

Then you need to use the address-of operator & on the variable to get a pointer to the variable, and cast that pointer:

foo(reinterpret_cast<LPBYTE>(&y));

---

If the variable actually hold an address, then you first of all have to be very careful because it's not guaranteed that int can hold a memory address (ie a pointer). Think for example on a 64-bit system where pointers are 64 bits, but int is usually still a 32 bit type.

Use instead intptr_t which is guaranteed to be big enough to hold either an int or a pointer.

Then you should do eg

 
 
 
  
  intptr_t y = ...; foo(reinterpret_cast<LPBYTE>(y));
 
  

Answer 2

I'm not sure what is far word says.

far is a 32-bit pointer which consists of 16-bit 'selector' that helps in determining the start address of the memory region, and a 16-bit offset into the memory region.

You can cast your variable like this:

int main()
{
 int y = koo();
 foo(static_cast<LPBYTE>(&y)); 
 return 0;
}