理解GCC内联汇编语法中的输入/输出操作数

Question

As part of writing my OS, I am implementing interrupt handling and the I/O functions inb and outb .

I had to learn writing inline assembly in GCC and read up a lot about it online. Based on my understanding, I wrote my own code. Meanwhile, I looked up Linux's implementation of the functions from /usr/include/sys/io.h . This is what it is like for outb :

static __inline void
outb (unsigned char __value, unsigned short int __port)
{
  __asm__ __volatile__ ("outb %b0,%w1": :"a" (__value), "Nd" (__port));
}

Here are my questions:

• The GCC manual says "N" is

Unsigned 8-bit integer constant (for in and out instructions).

But here __port is unsigned short int which I believe would be 16 bits. So how is it decided which portion of the 16 bits is used in the inline assembly ?

• This is my understanding of how this works - value of __port will be used directly (because of the "N") as a constant in place of %w1. Value of __value is copied to eax . %bo is replaced by %al. Then the instruction is executed. Is this correct ?

• How is it decided which of "N" or "d" to use for the second operand ? Is there some preference order ?

• What difference does it make if I don't use "N" ? Wouldn't simply using "d" be better, since that is 16 bits ?

• If I omitted the "N", then is it correct that value of __port is copied to edx and then %w1 is replaced by edx ?

Answer 1

I may be mistaken, but my understanding is that "Nd" means use either N or d , at the compiler's preference. If the value is not known to be a constant that fits in 8 bits, then N is not satisfiable, so the compiler has no choice but to use d . But when the value is a compile-time constant and its value fits in 8 bits, using an 8-bit immediate is preferable to wasting a register.