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XOR the orig and result bit 7 for overflow

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We should have done this all along. We hadn't accounted for if orig >=
0x80 and result < 0x80, but at that point, it's kind of silly to have a
long complicated condition that we can satisfy simply with an XOR.
(Which is what the machine would do anyway.)
This commit is contained in:
Peter Evans 2018-01-21 00:05:49 -06:00
parent 809e6ca7c6
commit 42c7fcbb47
1 changed files with 6 additions and 1 deletions
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7
src/mos6502.c
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@ -232,6 +232,8 @@ mos6502_set_status(mos6502 *cpu, vm_8bit status)
void
mos6502_modify_status(mos6502 *cpu, vm_8bit status, int orig, int result)
{
int bit7o, bit7r;
if (status & MOS_NEGATIVE) {
cpu->P &= ~MOS_NEGATIVE;
if (result & 0x80) {
@ -242,12 +244,15 @@ mos6502_modify_status(mos6502 *cpu, vm_8bit status, int orig, int result)
if (status & MOS_OVERFLOW) {
cpu->P &= ~MOS_OVERFLOW;
bit7o = orig & 0x80;
bit7r = result & 0x80;
// If the result of the operation is such that the sign bit,
// that is to say bit 7, changes, then we have overflowed. E.g.:
// 90 + 40 = 130, but that's actually -124 in two's complement.
// So if you are paying attention to the sign, you have
// overflowed from a positive into a negative result.
if (orig < 0x80 && result >= 0x80) {
if (bit7o ^ bit7r) {
cpu->P |= MOS_OVERFLOW;
}
}