解释AF标志在x86指令是如何工作的？

问题描述

我有一点点的8086仿真器和我已经像一个长期的错误2年现在AF无法正常运行子内，并添加说明。

我现在计算其价值的方式是这样的8位数字和加减：

  uint8_t有基地= ...，SUBT = ...
基地=基本和放大器; 0xF的;
SUBT = SUBT＆放大器; 0xF的; //隔离底部蚕食
如果（（int16_t）基SUBT大于7 ||（int16_t）基SUBT＆下; -7）{
    flags.af = 1;
}其他{
   flags.af = 0;
}
 

（假设像子基地的指令，SUBT ）

和添加是这样的：

  uint8_t有基地= ...，加法器= ...
基地=基本和放大器; 0xF的;
加法=加法器与放大器; 0xF的; //隔离底部蚕食
如果（基地+加法大于7 ||基地+加法＆LT; -7）{
    flags.af = 1;
}其他{
   flags.af = 0;
}
 

（像指令添加基地，加法器）

我如何正确计算AF标志在我的模拟器这样的指令？

解决方案

  flags.af =（（（基SUBT）及〜0xF的）！= 0）;
 

检查是否上部比特是除零任何东西，这将指示上溢或下溢出在底面4位

下面是一个版本，这是一个有点接近原件。注意，两个4位的量之间的差值不会比15更大同样地，除了将永远不会小于0

  flags.af =（（int16_t）基SUBT＆LT; -15）;flags.af =（（int16_t）基地+加法器→15）;
 

把周围一个布尔前pression括号是只是一种风格$矿的对$ pference，我知道他们是多余的。

I have a little 8086 emulator and I've had a long standing bug for like 2 years now that AF does not behave properly inside of sub and add instructions.

My current way of computing its value is this for 8 bit numbers and subtraction:

uint8_t base=... , subt=...
base=base&0xF;
subt=subt&0xF; //isolate bottom nibble
if((int16_t)base-subt>7 || (int16_t)base-subt<-7){
    flags.af=1;
}else{
   flags.af=0;
}

(assuming an instruction like sub base,subt )

And for adding it's like this:

uint8_t base=... , adder=...
base=base&0xF;
adder=adder&0xF; //isolate bottom nibble
if(base+adder>7 || base+adder<-7){
    flags.af=1;
}else{
   flags.af=0;
}

(for an instruction like add base,adder)

How do I properly calculate the AF flag in my emulator for such instructions?

解决方案

flags.af = (((base-subt) & ~0xf) != 0);

Checks to see if the upper bits are anything except zero, which would indicate an overflow or underflow out of the bottom 4 bits.

Here's a version that's a little closer to your original. Note that the difference between two 4-bit quantities will never be greater than 15. Likewise the addition will never be less than 0.

flags.af = ((int16_t)base-subt < -15);

flags.af = ((int16_t)base+adder > 15);

Putting parentheses around a boolean expression is just a style preference of mine, I know they're redundant.

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