Hi,
On 16.08.2017 05:17, 王翔 wrote:
The source code may have a problem when the IDT is initialized. This code is located in `src/cpu/x86/16bit/entry16.inc`.
movw %cs, %ax shlw $4, %ax movw $nullidt_offset, %bx subw %ax, %bx lidt %cs:(%bx) —> this mybe lidtl movw $gdtptr16_offset, %bx subw %ax, %bx lgdtl %cs:(%bx)
lidt only used for base 0-16M. But nullidt located in (4G-64k) - 4G space.
I might be wrong here because I'm reading this code for the first time. But it looks to me as if `nullidt` is not an IDT but just a null pointer (and null limit) to zero the IDTR register (i.e. SRC in the description of LIDT is the zero values below `nullidt:` in the source).
Nico
This instruct must be **lidtl**
Description of LIDT IF Instruction is LIDT THEN IF OperandSize = 16 THEN IDTR(Limit) ← SRC[0:15]; IDTR(Base) ← SRC[16:47] AND 00FFFFFFH; ELSE IF 32-bit Operand Size THEN IDTR(Limit) ← SRC[0:15]; IDTR(Base) ← SRC[16:47]; FI; ELSE IF 64-bit Operand Size (* In 64-Bit Mode *) THEN IDTR(Limit) ← SRC[0:15]; IDTR(Base) ← SRC[16:79]; FI; FI; ELSE (* Instruction is LGDT *) IF OperandSize = 16 THEN GDTR(Limit) ← SRC[0:15]; GDTR(Base) ← SRC[16:47] AND 00FFFFFFH; ELSE IF 32-bit Operand Size THEN GDTR(Limit) ← SRC[0:15]; GDTR(Base) ← SRC[16:47]; FI; ELSE IF 64-bit Operand Size (* In 64-Bit Mode *) THEN GDTR(Limit) ← SRC[0:15]; GDTR(Base) ← SRC[16:79]; FI; FI; FI;