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Updated SM2 MFMM (2) (markdown)
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@ -97,6 +97,8 @@ $t_2=t_2 - a_1$
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$t_3=t_3 - a_0$
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$t_0=t_0 - a_1$
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**减法显然是安全的(因为第四步的结果显然是>=0的,而且为零的情况仅限于 $t_0==0$ 的情况 ),所以调整为先做减法,再做加法,确保第四步加法不会产生进位。**。
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伪代码:
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```asm
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\ // First reduction step, [p3, p2, p1, p0] = [1, -0x100000000, 0, (1 - 0x100000000), -1]
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@ -104,17 +106,17 @@ $t_0=t_0 - a_1$
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MOVQ acc0, DX \
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SHLQ $32, AX \ // AX = L(acc0 * 2^32), low part
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SHRQ $32, DX \ // DX = H(acc0 * 2^32), high part
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\ // calculate the positive part first: [1, 0, 0, 1] * acc0 + [0, acc3, acc2, acc1],
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\ // due to (-1) * acc0 + acc0 == 0, so last lowest lamb 0 is dropped directly, no carry.
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ADDQ acc0, acc1 \ // acc1' = L (acc0 + acc1)
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ADCQ $0, acc2 \ // acc2' = acc2 + carry1
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ADCQ $0, acc3 \ // acc3' = acc3 + carry2
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ADCQ $0, acc0 \ // acc0' = acc0 + carry3
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\// calculate the negative part: [0, -0x100000000, 0, -0x100000000] * acc0
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SUBQ AX, acc1 \
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SBBQ DX, acc2 \
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SBBQ AX, acc3 \
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SBBQ DX, acc0 \
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\ // calculate the positive part: [1, 0, 0, 1] * acc0 + [0, acc3, acc2, acc1],
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\ // due to (-1) * acc0 + acc0 == 0, so last lowest lamb 0 is dropped directly, no carry.
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ADDQ acc0, acc1 \ // acc1' = L (acc0 + acc1)
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ADCQ $0, acc2 \ // acc2' = acc2 + carry1
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ADCQ $0, acc3 \ // acc3' = acc3 + carry2
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ADCQ $0, acc0 \ // acc0' = acc0 + carry3
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```
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移位: 2
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加法:4
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