溢出和Z80携带旗帜 [英] Overflow and Carry flags on Z80
问题描述
我已经得到了全面实施的ADD A，我的Z80核心研发集运codeS的。我有一些关于进位和溢出标志，我想我已经钉混乱，但我希望把它放到社区来检查我是对的。
I have gotten round to implementing the ADD A,r set of opcodes on my Z80 core. I had a bit of confusion about the carry and overflow flags which I think I've nailed, but I wanted to put it to the community to check that I'm right.
基本上，从我所看到的，ALU在Z80并不关心符号/无符号运算，它只是增加了位。这意味着，如果两个8位值加在一起，并导致一个9位值作为其相加结果，进位标志将被设置。这包括添加两个负二的补数，例如-20（11101100）和-40（11011000），作为虽然结果是-60（11000100），结果实际上是一个9位值1 1100 0100这确实意味着如果将两个负二的补值，进位标志将始终置，即使在没有溢出的情况 - 我说的对
Basically, from what I can see, the ALU in the Z80 doesn't care about signed/unsigned operations, it just adds bits. This means that if two 8-bit values are added together and cause a 9-bit value as a result of their addition, the carry flag will be set. This includes adding two negative two's complement numbers, for example -20 (11101100) and -40 (11011000), as although the result is -60 (11000100), the result is actually a 9-bit value 1 1100 0100. This surely means if adding two negative two's complement values, the carry flag will always be set, even when there is no overflow condition - am I right?
其次，我决定，在该指令检测溢出，我会XOR位两个操作数的7，如果结果是为10000000，那么肯定是没有溢出 - 如果这样做的结果是00000000则有可能是溢出的迹象是相同的，因此，我会XOR位7加用一个操作数的第7位的结果，如果这样做的结果是千万然后溢出已经发生了，我设置了P / V溢出旗。难道我在这里也？
Secondly, I decided that to detect an overflow in this instruction, I would XOR bit 7 of both operands, and if the result was 10000000, then there is definitely no overflow - if the result of this is 00000000 then there could be an overflow as the signs are the same, and I would therefore XOR bit 7 of the result of the addition with bit 7 of either operand, and if the result of this is 10000000 then an overflow has occurred and I set the P/V overflow flag. Am I right here also?
对不起，这样一个令人费解的问题，我是pretty肯定我的权利，但我需要知道之前，我进行基于这个逻辑无数更多的指令。非常感谢。
Sorry for such a convoluted question, I'm pretty sure I'm right but I need to know before I carry on with countless more instructions based on this logic. Many thanks.
推荐答案
结果的位从无符号整数的截断和获得。 ADD指令并不关心符号在这里也不关心整数自己际pretation为带符号。它只是增加了好像数字是无符号的。
The bits of the result are obtained from the truncated sum of unsigned integers. The add instruction doesn't care about the sign here nor does it care about your own interpretation of the integers as signed or unsigned. It just adds as if the numbers were unsigned.
，进位标记（或借在减法的情况下）是从相加的8位无符号整数的不存在的第9位。实际上，这意味着标记为无符号整数添加/分的溢/下溢。此外，加不关心这里的招牌可言，它只是增加了好像数字是无符号的。
The carry flag (or borrow in case of subtraction) is that non-existent 9th bit from the addition of the 8-bit unsigned integers. Effectively, this flag signifies an overflow/underflow for add/sub of unsigned integers. Again, add doesn't care about the signs here at all, it just adds as if the numbers were unsigned.
添加两负2的补数将导致进位标志设置为1，正确的。
Adding two negative 2's complement numbers will result in setting of the carry flag to 1, correct.
溢出标志显示是否有一直为有符号整数添加/分的溢/下溢。要设置溢出标志指令对待数字作为符号（就像它会将它们视为无符号的进位标志和结果的8位）。
The overflow flag shows whether or not there's been an overflow/underflow for add/sub of signed integers. To set the overflow flag the instruction treats the numbers as signed (just like it treats them as unsigned for the carry flag and the 8 bits of the result).
背后设置溢出标志的想法很简单。假设你的8位有符号整数符号扩展到9位，也就是说，仅仅是第7位复制到一个额外的，第8位。如果这些9位带符号整数的9位和/差具有在比特7和8个不同的值，这意味着该加/减已经失去了在第7位的结果的标志，并用它在将发生上溢/下溢结果的数量级，或者，换句话说，8位数据不能容纳符号位和这样大的幅度。
The idea behind setting the overflow flag is simple. Suppose you sign-extend your 8-bit signed integers to 9 bits, that is, just copy the 7th bit to an extra, 8th bit. An overflow/underflow will occur if the 9-bit sum/difference of these 9-bit signed integers has different values in bits 7 and 8, meaning that the addition/subtraction has lost the result's sign in the 7th bit and used it for the result's magnitude, or, in other words, the 8 bits can't accommodate the sign bit and such a large magnitude.
现在，位结果的7可从虚符号位8不同，当且仅当进位位7和进位位的8（=进行7位的）是不同的。这是因为我们开始与加数有7位= 8位，只携带不同的插件到他们可以影响他们的结果以不同的方式。
Now, bit 7 of the result can differ from the imaginary sign bit 8 if and only if the carry into bit 7 and the carry into bit 8 (=carry out of bit 7) are different. That's because we start with the addends having bit 7=bit 8 and only different carry-ins into them can affect them in the result in different ways.
所以溢出标志=进位输出，从第6位标志XOR进位第7位。
So overflow flag = carry-out flag XOR carry from bit 6 into bit 7.
我都和你计算溢出标志的方式是正确的。事实上，无论是在 Z80 CPU用户手册节Z80状态指示标志中所述。
Both my and your ways of calculating the overflow flag are correct. In fact, both are described in the Z80 CPU User's Manual in section "Z80 Status Indicator Flags".
下面是你如何能模仿最C，对ADC指令，你不必CPU的标志直接进入，不能充分利用CPU仿真的ADC指令的：
Here's how you can emulate most of the ADC instruction in C, where you don't have direct access to the CPU's flags and can't take full advantage of the emulating CPU's ADC instruction:
#include <stdio.h>
#include <limits.h>
#if CHAR_BIT != 8
#error char expected to have exactly 8 bits.
#endif
typedef unsigned char uint8;
typedef signed char int8;
#define FLAGS_CY_SHIFT 0
#define FLAGS_OV_SHIFT 1
#define FLAGS_CY_MASK (1 << FLAGS_CY_SHIFT)
#define FLAGS_OV_MASK (1 << FLAGS_OV_SHIFT)
void Adc(uint8* acc, uint8 b, uint8* flags)
{
uint8 a = *acc;
uint8 carryIns;
uint8 carryOut;
// Calculate the carry-out depending on the carry-in and addends.
//
// carry-in = 0: carry-out = 1 IFF (a + b > 0xFF) or,
// equivalently, but avoiding overflow in C: (a > 0xFF - b).
//
// carry-in = 1: carry-out = 1 IFF (a + b + 1 > 0xFF) or,
// equivalently, (a + b >= 0xFF) or,
// equivalently, but avoiding overflow in C: (a >= 0xFF - b).
//
// Also calculate the sum bits.
if (*flags & FLAGS_CY_MASK)
{
carryOut = (a >= 0xFF - b);
*acc = a + b + 1;
}
else
{
carryOut = (a > 0xFF - b);
*acc = a + b;
}
#if 0
// Calculate the overflow by sign comparison.
carryIns = ((a ^ b) ^ 0x80) & 0x80;
if (carryIns) // if addend signs are different
{
// overflow if the sum sign differs from the sign of either of addends
carryIns = ((*acc ^ a) & 0x80) != 0;
}
#else
// Calculate all carry-ins.
// Remembering that each bit of the sum =
// addend a's bit XOR addend b's bit XOR carry-in,
// we can work out all carry-ins from a, b and their sum.
carryIns = *acc ^ a ^ b;
// Calculate the overflow using the carry-out and
// most significant carry-in.
carryIns = (carryIns >> 7) ^ carryOut;
#endif
// Update flags.
*flags &= ~(FLAGS_CY_MASK | FLAGS_OV_MASK);
*flags |= (carryOut << FLAGS_CY_SHIFT) | (carryIns << FLAGS_OV_SHIFT);
}
void Sbb(uint8* acc, uint8 b, uint8* flags)
{
// a - b - c = a + ~b + 1 - c = a + ~b + !c
*flags ^= FLAGS_CY_MASK;
Adc(acc, ~b, flags);
*flags ^= FLAGS_CY_MASK;
}
const uint8 testData[] =
{
0,
1,
0x7F,
0x80,
0x81,
0xFF
};
int main(void)
{
unsigned aidx, bidx, c;
printf("ADC:\n");
for (c = 0; c <= 1; c++)
for (aidx = 0; aidx < sizeof(testData)/sizeof(testData[0]); aidx++)
for (bidx = 0; bidx < sizeof(testData)/sizeof(testData[0]); bidx++)
{
uint8 a = testData[aidx];
uint8 b = testData[bidx];
uint8 flags = c << FLAGS_CY_SHIFT;
printf("%3d(%4d) + %3d(%4d) + %u = ",
a, (int8)a, b, (int8)b, c);
Adc(&a, b, &flags);
printf("%3d(%4d) CY=%d OV=%d\n",
a, (int8)a, (flags & FLAGS_CY_MASK) != 0, (flags & FLAGS_OV_MASK) != 0);
}
printf("SBB:\n");
for (c = 0; c <= 1; c++)
for (aidx = 0; aidx < sizeof(testData)/sizeof(testData[0]); aidx++)
for (bidx = 0; bidx < sizeof(testData)/sizeof(testData[0]); bidx++)
{
uint8 a = testData[aidx];
uint8 b = testData[bidx];
uint8 flags = c << FLAGS_CY_SHIFT;
printf("%3d(%4d) - %3d(%4d) - %u = ",
a, (int8)a, b, (int8)b, c);
Sbb(&a, b, &flags);
printf("%3d(%4d) CY=%d OV=%d\n",
a, (int8)a, (flags & FLAGS_CY_MASK) != 0, (flags & FLAGS_OV_MASK) != 0);
}
return 0;
}
输出：
ADC:
0( 0) + 0( 0) + 0 = 0( 0) CY=0 OV=0
0( 0) + 1( 1) + 0 = 1( 1) CY=0 OV=0
0( 0) + 127( 127) + 0 = 127( 127) CY=0 OV=0
0( 0) + 128(-128) + 0 = 128(-128) CY=0 OV=0
0( 0) + 129(-127) + 0 = 129(-127) CY=0 OV=0
0( 0) + 255( -1) + 0 = 255( -1) CY=0 OV=0
1( 1) + 0( 0) + 0 = 1( 1) CY=0 OV=0
1( 1) + 1( 1) + 0 = 2( 2) CY=0 OV=0
1( 1) + 127( 127) + 0 = 128(-128) CY=0 OV=1
1( 1) + 128(-128) + 0 = 129(-127) CY=0 OV=0
1( 1) + 129(-127) + 0 = 130(-126) CY=0 OV=0
1( 1) + 255( -1) + 0 = 0( 0) CY=1 OV=0
127( 127) + 0( 0) + 0 = 127( 127) CY=0 OV=0
127( 127) + 1( 1) + 0 = 128(-128) CY=0 OV=1
127( 127) + 127( 127) + 0 = 254( -2) CY=0 OV=1
127( 127) + 128(-128) + 0 = 255( -1) CY=0 OV=0
127( 127) + 129(-127) + 0 = 0( 0) CY=1 OV=0
127( 127) + 255( -1) + 0 = 126( 126) CY=1 OV=0
128(-128) + 0( 0) + 0 = 128(-128) CY=0 OV=0
128(-128) + 1( 1) + 0 = 129(-127) CY=0 OV=0
128(-128) + 127( 127) + 0 = 255( -1) CY=0 OV=0
128(-128) + 128(-128) + 0 = 0( 0) CY=1 OV=1
128(-128) + 129(-127) + 0 = 1( 1) CY=1 OV=1
128(-128) + 255( -1) + 0 = 127( 127) CY=1 OV=1
129(-127) + 0( 0) + 0 = 129(-127) CY=0 OV=0
129(-127) + 1( 1) + 0 = 130(-126) CY=0 OV=0
129(-127) + 127( 127) + 0 = 0( 0) CY=1 OV=0
129(-127) + 128(-128) + 0 = 1( 1) CY=1 OV=1
129(-127) + 129(-127) + 0 = 2( 2) CY=1 OV=1
129(-127) + 255( -1) + 0 = 128(-128) CY=1 OV=0
255( -1) + 0( 0) + 0 = 255( -1) CY=0 OV=0
255( -1) + 1( 1) + 0 = 0( 0) CY=1 OV=0
255( -1) + 127( 127) + 0 = 126( 126) CY=1 OV=0
255( -1) + 128(-128) + 0 = 127( 127) CY=1 OV=1
255( -1) + 129(-127) + 0 = 128(-128) CY=1 OV=0
255( -1) + 255( -1) + 0 = 254( -2) CY=1 OV=0
0( 0) + 0( 0) + 1 = 1( 1) CY=0 OV=0
0( 0) + 1( 1) + 1 = 2( 2) CY=0 OV=0
0( 0) + 127( 127) + 1 = 128(-128) CY=0 OV=1
0( 0) + 128(-128) + 1 = 129(-127) CY=0 OV=0
0( 0) + 129(-127) + 1 = 130(-126) CY=0 OV=0
0( 0) + 255( -1) + 1 = 0( 0) CY=1 OV=0
1( 1) + 0( 0) + 1 = 2( 2) CY=0 OV=0
1( 1) + 1( 1) + 1 = 3( 3) CY=0 OV=0
1( 1) + 127( 127) + 1 = 129(-127) CY=0 OV=1
1( 1) + 128(-128) + 1 = 130(-126) CY=0 OV=0
1( 1) + 129(-127) + 1 = 131(-125) CY=0 OV=0
1( 1) + 255( -1) + 1 = 1( 1) CY=1 OV=0
127( 127) + 0( 0) + 1 = 128(-128) CY=0 OV=1
127( 127) + 1( 1) + 1 = 129(-127) CY=0 OV=1
127( 127) + 127( 127) + 1 = 255( -1) CY=0 OV=1
127( 127) + 128(-128) + 1 = 0( 0) CY=1 OV=0
127( 127) + 129(-127) + 1 = 1( 1) CY=1 OV=0
127( 127) + 255( -1) + 1 = 127( 127) CY=1 OV=0
128(-128) + 0( 0) + 1 = 129(-127) CY=0 OV=0
128(-128) + 1( 1) + 1 = 130(-126) CY=0 OV=0
128(-128) + 127( 127) + 1 = 0( 0) CY=1 OV=0
128(-128) + 128(-128) + 1 = 1( 1) CY=1 OV=1
128(-128) + 129(-127) + 1 = 2( 2) CY=1 OV=1
128(-128) + 255( -1) + 1 = 128(-128) CY=1 OV=0
129(-127) + 0( 0) + 1 = 130(-126) CY=0 OV=0
129(-127) + 1( 1) + 1 = 131(-125) CY=0 OV=0
129(-127) + 127( 127) + 1 = 1( 1) CY=1 OV=0
129(-127) + 128(-128) + 1 = 2( 2) CY=1 OV=1
129(-127) + 129(-127) + 1 = 3( 3) CY=1 OV=1
129(-127) + 255( -1) + 1 = 129(-127) CY=1 OV=0
255( -1) + 0( 0) + 1 = 0( 0) CY=1 OV=0
255( -1) + 1( 1) + 1 = 1( 1) CY=1 OV=0
255( -1) + 127( 127) + 1 = 127( 127) CY=1 OV=0
255( -1) + 128(-128) + 1 = 128(-128) CY=1 OV=0
255( -1) + 129(-127) + 1 = 129(-127) CY=1 OV=0
255( -1) + 255( -1) + 1 = 255( -1) CY=1 OV=0
SBB:
0( 0) - 0( 0) - 0 = 0( 0) CY=0 OV=0
0( 0) - 1( 1) - 0 = 255( -1) CY=1 OV=0
0( 0) - 127( 127) - 0 = 129(-127) CY=1 OV=0
0( 0) - 128(-128) - 0 = 128(-128) CY=1 OV=1
0( 0) - 129(-127) - 0 = 127( 127) CY=1 OV=0
0( 0) - 255( -1) - 0 = 1( 1) CY=1 OV=0
1( 1) - 0( 0) - 0 = 1( 1) CY=0 OV=0
1( 1) - 1( 1) - 0 = 0( 0) CY=0 OV=0
1( 1) - 127( 127) - 0 = 130(-126) CY=1 OV=0
1( 1) - 128(-128) - 0 = 129(-127) CY=1 OV=1
1( 1) - 129(-127) - 0 = 128(-128) CY=1 OV=1
1( 1) - 255( -1) - 0 = 2( 2) CY=1 OV=0
127( 127) - 0( 0) - 0 = 127( 127) CY=0 OV=0
127( 127) - 1( 1) - 0 = 126( 126) CY=0 OV=0
127( 127) - 127( 127) - 0 = 0( 0) CY=0 OV=0
127( 127) - 128(-128) - 0 = 255( -1) CY=1 OV=1
127( 127) - 129(-127) - 0 = 254( -2) CY=1 OV=1
127( 127) - 255( -1) - 0 = 128(-128) CY=1 OV=1
128(-128) - 0( 0) - 0 = 128(-128) CY=0 OV=0
128(-128) - 1( 1) - 0 = 127( 127) CY=0 OV=1
128(-128) - 127( 127) - 0 = 1( 1) CY=0 OV=1
128(-128) - 128(-128) - 0 = 0( 0) CY=0 OV=0
128(-128) - 129(-127) - 0 = 255( -1) CY=1 OV=0
128(-128) - 255( -1) - 0 = 129(-127) CY=1 OV=0
129(-127) - 0( 0) - 0 = 129(-127) CY=0 OV=0
129(-127) - 1( 1) - 0 = 128(-128) CY=0 OV=0
129(-127) - 127( 127) - 0 = 2( 2) CY=0 OV=1
129(-127) - 128(-128) - 0 = 1( 1) CY=0 OV=0
129(-127) - 129(-127) - 0 = 0( 0) CY=0 OV=0
129(-127) - 255( -1) - 0 = 130(-126) CY=1 OV=0
255( -1) - 0( 0) - 0 = 255( -1) CY=0 OV=0
255( -1) - 1( 1) - 0 = 254( -2) CY=0 OV=0
255( -1) - 127( 127) - 0 = 128(-128) CY=0 OV=0
255( -1) - 128(-128) - 0 = 127( 127) CY=0 OV=0
255( -1) - 129(-127) - 0 = 126( 126) CY=0 OV=0
255( -1) - 255( -1) - 0 = 0( 0) CY=0 OV=0
0( 0) - 0( 0) - 1 = 255( -1) CY=1 OV=0
0( 0) - 1( 1) - 1 = 254( -2) CY=1 OV=0
0( 0) - 127( 127) - 1 = 128(-128) CY=1 OV=0
0( 0) - 128(-128) - 1 = 127( 127) CY=1 OV=0
0( 0) - 129(-127) - 1 = 126( 126) CY=1 OV=0
0( 0) - 255( -1) - 1 = 0( 0) CY=1 OV=0
1( 1) - 0( 0) - 1 = 0( 0) CY=0 OV=0
1( 1) - 1( 1) - 1 = 255( -1) CY=1 OV=0
1( 1) - 127( 127) - 1 = 129(-127) CY=1 OV=0
1( 1) - 128(-128) - 1 = 128(-128) CY=1 OV=1
1( 1) - 129(-127) - 1 = 127( 127) CY=1 OV=0
1( 1) - 255( -1) - 1 = 1( 1) CY=1 OV=0
127( 127) - 0( 0) - 1 = 126( 126) CY=0 OV=0
127( 127) - 1( 1) - 1 = 125( 125) CY=0 OV=0
127( 127) - 127( 127) - 1 = 255( -1) CY=1 OV=0
127( 127) - 128(-128) - 1 = 254( -2) CY=1 OV=1
127( 127) - 129(-127) - 1 = 253( -3) CY=1 OV=1
127( 127) - 255( -1) - 1 = 127( 127) CY=1 OV=0
128(-128) - 0( 0) - 1 = 127( 127) CY=0 OV=1
128(-128) - 1( 1) - 1 = 126( 126) CY=0 OV=1
128(-128) - 127( 127) - 1 = 0( 0) CY=0 OV=1
128(-128) - 128(-128) - 1 = 255( -1) CY=1 OV=0
128(-128) - 129(-127) - 1 = 254( -2) CY=1 OV=0
128(-128) - 255( -1) - 1 = 128(-128) CY=1 OV=0
129(-127) - 0( 0) - 1 = 128(-128) CY=0 OV=0
129(-127) - 1( 1) - 1 = 127( 127) CY=0 OV=1
129(-127) - 127( 127) - 1 = 1( 1) CY=0 OV=1
129(-127) - 128(-128) - 1 = 0( 0) CY=0 OV=0
129(-127) - 129(-127) - 1 = 255( -1) CY=1 OV=0
129(-127) - 255( -1) - 1 = 129(-127) CY=1 OV=0
255( -1) - 0( 0) - 1 = 254( -2) CY=0 OV=0
255( -1) - 1( 1) - 1 = 253( -3) CY=0 OV=0
255( -1) - 127( 127) - 1 = 127( 127) CY=0 OV=1
255( -1) - 128(-128) - 1 = 126( 126) CY=0 OV=0
255( -1) - 129(-127) - 1 = 125( 125) CY=0 OV=0
255( -1) - 255( -1) - 1 = 255( -1) CY=1 OV=0
您可以修改 0＃如果到＃如果1 使用溢出基于符号比较方法计算。结果将是相同的。乍一看这是一个有点令人惊讶的是基于登录方法将随身携带的照顾了。
You can change #if 0 to #if 1 to use the sign-comparison-based method for overflow calculation. The result will be the same. At first glance it's a bit surprising that the sign-based method takes care of the carry-in too.
请注意，用我的方法中，我通过7计算所有携带的插件为位0，你也得到半进标志自由的价值（来自第3位至第4位携带）这就是所需 DAA 指令。
Please note that by using my method in which I calculate all carry-ins into bits 0 through 7, you also get for free the value of the half-carry flag (carry from bit 3 to bit 4) that's needed for the DAA instruction.
编辑：我添加了一个功能，可借（SBC / SBB指令）和结果它减法
I've added a function for subtraction with borrow (SBC/SBB instruction) and results for it.
这篇关于溢出和Z80携带旗帜的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
