struct task_struct * get_current(void)函数的解释--help!!

[arcsiny]

static inline struct task_struct * get_current(void)
{
struct task_struct *current;
__asm__("andl %%esp,%0; ":"=r" (current) : "0" (~8191UL));
return current;
}
堆栈指针寄存器esp与“~8191UL"相与可以得到当前进程的起始地址。
请问esp里是什么内容？ ~8191UL 如何解释？UL?
谢谢！

[oldmaize]

get_current() is a routine for getting access to the task_struct of the currently executing task. It uses the often confusing inline assembly features of GCC to perform this, as follows :

| __asm__(

This signifies a piece of inline assembly that the compiler must insert into its output code. The __asm__ is the same as asm, but can't be disabled by command line flags.

| "andl %%esp,%0

"%%" is a macro that expands to a "%".
"%0" is a macro that expands to the first input/output specification.

So in this case, it takes the stack pointer (register %esp) and ANDs it into a register that contains 0xFFFFE000, leaving the result in that register.

Basically, the task's task_struct and a task's kernel stack occupy an 8KB block that is 8KB aligned, with the task_struct at the beginning and the stack growing from the end downwards. So you can find the task_struct by clearing the bottom 13 bits of the stack pointer value.

| ; "

The semicolon can be used to separate assembly statements, as can the newline character escape sequence ("\n").

| :"=r" (current)

This specifies an output constraint (all of which occur after the first colon, but before the second). The '=' also specifies that this is an output. The 'r' indicates that a general purpose register should be allocated such that the instruction can place the output value into it. The bit inside the brackets - 'current' - is the intended destination of the output value (normally a local variable) once the C part is returned to.

| : "0" (~8191UL));

This specifies an input constraint (all of which occur after the second colon, but before the third). The '0' references another constraint (in this case, the first output constraint), saying that the same register or memory location should be used for both. The '~8191UL' inside the brackets is a constant that should be loaded into the register allocated for the output value before using the instructions inside the asm block.

See also the gcc info pages, Topic "C Extensions", subtopic "Extended Asm".

[oldmaize]

From
kernel 2.2 when a process is created its task_struct and its kernel
mode stack are placed together in two contiguos page frames (so 8KB).
They are organized "with the task_struct at the beginning and the
stack growing from the end downwards" as you correctly said. Now when
we want to obtain task descriptor (address of task_struct) we use
%esp. Why? Because if you're in kernel mode %esp is the kernel mode
stack pointer. Now you can easily notice that zeroing the last %esp 13
bytes gives you exactly what you need since it will point to the
beginning of the first page frame (where task_struct is).
Now, with this in mind, it's easy to understand get_current().

This is inline asm syntax.

asm ( assembler template
: output operands (optional)
: input operands (optional)
: list of clobbered registers (optional)
);

When you see "0" in input operands in get_current it means it's the
same of first argument. "=r" stands for register and so the output
operand is to be stored in a register. Now get_current() should appear
simple. It loads ~8191UL in current and then ANDs it with %esp
storing the result in current.

[arcsiny]

多谢了
基本明白一点
不过还有 UL 是什么意思？