mirror of
https://github.com/TomHarte/CLK.git
synced 2024-12-25 18:30:21 +00:00
Takes a second pass at DIVS timing, seeming to correct that side of things.
This commit is contained in:
parent
f88dc23c71
commit
e3794c0c0e
@ -1049,22 +1049,28 @@ template <class T, bool dtack_is_implicit, bool signal_will_perform> void Proces
|
||||
break;
|
||||
}
|
||||
|
||||
int32_t dividend = int32_t(destination()->full);
|
||||
int32_t divisor = s_extend16(source()->halves.low.full);
|
||||
const int64_t quotient = int64_t(dividend) / int64_t(divisor);
|
||||
const int32_t signed_dividend = int32_t(destination()->full);
|
||||
const int32_t signed_divisor = s_extend16(source()->halves.low.full);
|
||||
const auto result_sign =
|
||||
( (0 <= signed_dividend) - (signed_dividend < 0) ) *
|
||||
( (0 <= signed_divisor) - (signed_divisor < 0) );
|
||||
|
||||
const uint32_t dividend = uint32_t(abs(signed_dividend));
|
||||
const uint32_t divisor = uint32_t(abs(signed_divisor));
|
||||
|
||||
int cycles_expended = 12; // Covers the nn nnn n to get beyond the sign test.
|
||||
if(dividend < 0) {
|
||||
if(signed_dividend < 0) {
|
||||
cycles_expended += 2; // An additional microycle applies if the dividend is negative.
|
||||
}
|
||||
|
||||
// Check for overflow. If it exists, work here is already done.
|
||||
if(quotient > 32767 || quotient < -32768) {
|
||||
const auto quotient = dividend / divisor;
|
||||
if(quotient > 32767) {
|
||||
overflow_flag_ = 1;
|
||||
set_next_microcycle_length(HalfCycles(3*2*2));
|
||||
set_next_microcycle_length(HalfCycles(6*2*2));
|
||||
|
||||
// These are officially undefined for results that overflow, so the below is a guess.
|
||||
zero_result_ = decltype(zero_result_)(divisor & 0xffff);
|
||||
zero_result_ = decltype(zero_result_)(dividend);
|
||||
negative_flag_ = zero_result_ & 0x8000;
|
||||
|
||||
break;
|
||||
@ -1074,26 +1080,25 @@ template <class T, bool dtack_is_implicit, bool signal_will_perform> void Proces
|
||||
negative_flag_ = zero_result_ & 0x8000;
|
||||
overflow_flag_ = 0;
|
||||
|
||||
// TODO: check sign rules here; am I necessarily giving the remainder the correct sign?
|
||||
// (and, if not, am I counting it in the correct direction?)
|
||||
const uint16_t remainder = uint16_t(dividend % divisor);
|
||||
const uint16_t remainder = uint16_t(signed_dividend % signed_divisor);
|
||||
const int signed_quotient = result_sign*int(quotient);
|
||||
destination()->halves.high.full = remainder;
|
||||
destination()->halves.low.full = uint16_t(quotient);
|
||||
destination()->halves.low.full = uint16_t(signed_quotient);
|
||||
|
||||
// Algorithm here: there is a fixed three-microcycle cost per bit set
|
||||
// in the unsigned quotient; there is an additional microcycle for
|
||||
// every bit that is set. Also, since the possibility of overflow
|
||||
// was already dealt with, it's now a smaller number.
|
||||
int positive_quotient_bits = int(abs(quotient)) & 0xfffe;
|
||||
// Algorithm here: there is a fixed cost per unset bit
|
||||
// in the first 15 bits of the unsigned quotient.
|
||||
auto positive_quotient_bits = ~quotient & 0xfffe;
|
||||
convert_to_bit_count_16(positive_quotient_bits);
|
||||
cycles_expended += 2 * positive_quotient_bits;
|
||||
|
||||
// There's then no way to terminate the loop that isn't at least six cycles long.
|
||||
cycles_expended += 6;
|
||||
// There's then no way to terminate the loop that isn't at least ten cycles long;
|
||||
// there's also a fixed overhead per bit. The two together add up to the 104 below.
|
||||
cycles_expended += 104;
|
||||
|
||||
if(divisor < 0) {
|
||||
// This picks up at 'No more bits' in yacht.txt's diagram.
|
||||
if(signed_divisor < 0) {
|
||||
cycles_expended += 2;
|
||||
} else if(dividend < 0) {
|
||||
} else if(signed_dividend < 0) {
|
||||
cycles_expended += 4;
|
||||
}
|
||||
set_next_microcycle_length(HalfCycles(cycles_expended * 2));
|
||||
|
Loading…
Reference in New Issue
Block a user