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Regarding edits by [[Special:Contributions/94.43.147.234|94.43.147.234]] ([[User talk:94.43.147.234|talk]]): Step 2 states:{{quote|text=If u and v are both even, then gcd(u, v) = 2·gcd(u/2, v/2), because 2 is a common divisor.}}
Therefore, before ''u'' and ''v'' may both be halved and ''shift'' incremented, the [[Least significant bit|LSB]] of ''u'' must be zero, '''and''' the LSB of ''v'' must be zero. The test (in line 12 of the C code fragment) would look like this (avoiding the short-circuit operators for simplicity):
for (shift = 0; (u & 1) == 0 & (v & 1) == 0; ++shift) {
By [[de Morgan's laws]], this can be rewritten as
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for (shift = 0; ((u | v) & 1) == 0; ++shift) {
As a further test, the code fragment as it stood (comprising <tt>((u & v) & 1) == 0</tt>) was
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