读取 CF、PF、ZF、SF、OF [英] Reading CF, PF, ZF, SF, OF

问题描述

我正在为自己的汇编语言编写一个虚拟机，我希望能够在执行加法等操作时设置 x86-64 架构中设置的进位、奇偶校验、零、符号和溢出标志.

I am writing a virtual machine for my own assembly language, I want to be able to set the carry, parity, zero, sign and overflowflags as they are set in the x86-64 architecture, when I perform operations such as addition.

注意事项:

• 我使用的是 Microsoft Visual C++ 2015 &英特尔 C++ 编译器 16.0
• 我正在编译为 Win64 应用程序.
• 我的虚拟机(目前)只对 8 位整数进行算术运算
• 我(目前)对任何其他标志(例如 AF)不感兴趣

我目前的解决方案是使用以下功能:

My current solution is using the following function:

void update_flags(uint16_t input)
{
    Registers::flags.carry = (input > UINT8_MAX);
    Registers::flags.zero = (input == 0);
    Registers::flags.sign = (input < 0);
    Registers::flags.overflow = (int16_t(input) > INT8_MAX || int16_t(input) < INT8_MIN);

    // I am assuming that overflow is handled by trunctation
    uint8_t input8 = uint8_t(input);
    // The parity flag
    int ones = 0;
    for (int i = 0; i < 8; ++i)
        if (input8 & (1 << i) != 0) ++ones;

    Registers::flags.parity = (ones % 2 == 0);
}

对于添加，我将使用如下:

Which for addition, I would use as follows:

uint8_t a, b;
update_flags(uint16_t(a) + uint16_t(b));
uint8_t c = a + b;

为了澄清，我想知道是否有更有效/更简洁的方法(例如直接访问 RFLAGS)另外我的代码可能不适用于其他操作(例如乘法)

To clarify, I want to know if there is a more efficient/neat way of doing this (such as by accessing RFLAGS directly) Also my code may not work for other operations (e.g. multiplication)

EDIT 2 我现在已将代码更新为:

EDIT 2 I have updated my code now to this:

void update_flags(uint32_t result)
{
    Registers::flags.carry = (result > UINT8_MAX);
    Registers::flags.zero = (result == 0);
    Registers::flags.sign = (int32_t(result) < 0);
    Registers::flags.overflow = (int32_t(result) > INT8_MAX || int32_t(result) < INT8_MIN);
    Registers::flags.parity = (_mm_popcnt_u32(uint8_t(result)) % 2 == 0);
}

还有一个问题，我的进位标志代码能正常工作吗?我还希望它为减法过程中发生的借用"设置正确.

One more question, will my code for the carry flag work properly?, I also want it to be set correctly for "borrows" that occur during subtraction.

注意:我正在虚拟化的汇编语言是我自己设计的，旨在简单并基于英特尔的 x86-64(即 Intel64)实现，因此我希望这些标志的行为方式大致相同.

Note: The assembly language I am virtualising is of my own design, meant to be simple and based of Intel's implementation of x86-64 (i.e. Intel64), and so I would like these flags to behave in mostly the same way.

推荐答案

我似乎已经解决了这个问题，方法是将更新标志的参数拆分为一个无符号和有符号的结果，如下所示:

I appear to have solved the problem, by splitting the arguments to update flags into an unsigned and signed result as follows:

void update_flags(int16_t unsigned_result, int16_t signed_result)
{
    Registers::flags.zero = unsigned_result == 0;
    Registers::flags.sign = signed_result < 0;
    Registers::flags.carry = unsigned_result < 0 || unsigned_result > UINT8_MAX;
    Registers::flags.overflow = signed_result < INT8_MIN || signed_result > INT8_MAX
}

对于加法(应该为有符号和无符号输入产生正确的结果)，我会执行以下操作:

For addition (which should produce the correct result for both signed & unsigned inputs) I would do the following:

int8_t a, b;
int16_t signed_result = int16_t(a) + int16_t(b);
int16_t unsigned_result = int16_t(uint8_t(a)) + int16_t(uint8_t(b));
update_flags(unsigned_result, signed_result);
int8_t c = a + b;

带符号的乘法我会做以下事情:

And signed multiplication I would do the following:

int8_t a, b;
int16_t result = int16_t(a) * int16_t(b);
update_flags(result, result);
int8_t c = a * b;

其他更新标志的操作依此类推

And so on for the other operations that update the flags

注意:我在这里假设 int16_t(a) 符号扩展，并且 int16_t(uint8_t(a)) 零扩展.

Note: I am assuming here that int16_t(a) sign extends, and int16_t(uint8_t(a)) zero extends.

我也决定不使用奇偶校验标志，如果我以后改变主意，我的 _mm_popcnt_u32 解决方案应该可以工作..

I have also decided against having a parity flag, my _mm_popcnt_u32 solution should work if I change my mind later..

附言感谢所有回复的人，这很有帮助.此外，如果有人能发现我的代码中的任何错误，我们将不胜感激.

P.S. Thank you to everyone who responded, it was very helpful. Also if anyone can spot any mistakes in my code, that would be appreciated.

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