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XOR the orig and result bit 7 for overflow
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.)
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@ -232,6 +232,8 @@ mos6502_set_status(mos6502 *cpu, vm_8bit status)
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void
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mos6502_modify_status(mos6502 *cpu, vm_8bit status, int orig, int result)
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{
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int bit7o, bit7r;
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if (status & MOS_NEGATIVE) {
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cpu->P &= ~MOS_NEGATIVE;
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if (result & 0x80) {
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@ -242,12 +244,15 @@ mos6502_modify_status(mos6502 *cpu, vm_8bit status, int orig, int result)
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if (status & MOS_OVERFLOW) {
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cpu->P &= ~MOS_OVERFLOW;
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bit7o = orig & 0x80;
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bit7r = result & 0x80;
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// If the result of the operation is such that the sign bit,
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// that is to say bit 7, changes, then we have overflowed. E.g.:
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// 90 + 40 = 130, but that's actually -124 in two's complement.
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// So if you are paying attention to the sign, you have
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// overflowed from a positive into a negative result.
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if (orig < 0x80 && result >= 0x80) {
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if (bit7o ^ bit7r) {
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cpu->P |= MOS_OVERFLOW;
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}
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}
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