Complicated

[James Parsons]

I have found this code and want to decompose it. I know what it does(Detect color/monochrome monitor) but i need to understand it

/* 
    Video card mono/colour detection.
    Return values:  false=colour
                    true=monochrome
*/
_Bool detectVideoType(void)
{
    char c=(*(volatile USHORT*)0x410)&0x30;
 
    //c can be 0x00 or 0x20 for colour, 0x30 for mono.
    return (c==0x30);
}

what are all the asterisks for and the volatile keyword what does it do

[Cubbi]

The volatile keyword indicates that an object so qualified is located at an address that is mapped to an I/O port (or some similar non-memory resource).

This fragment takes the number 0x410, which is apparently an MMIO port on whatever platform this was written for. Then it constructs a pointer with the type "pointer to a volatile USHORT" out of that address, using the cast expression (volatile USHORT*)0x410 (assuming USHORT means unsigned 16-bit type, this is equivalent to reinterpret_cast<volatile std::uint16_t*>(0x410) in C++)

Then, the program dereferences that pointer with the asterisk: this operation reads two bytes from the memory-mapped I/O ports 0x410 and 0x411.

Then it uses bitwise AND to pick out two bits (mask 0x30) from the result.

Then it stores the resulting value in a char, which truncates it to a single byte, which is then compared with one of the possible values.

PS: I see, this code fragment is from OSDev wiki and it's reading the BIOS data on a PC in real mode. volatile qualifier isn't strictly necessary, but works to be on the safe side. Either way, it's unlikely that you can encounter a platform where this test is necessary.