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B=5时 bucket num 由 hash 的低 5 位决定
// 比如 B=5,那 m 就是31,二进制是全 1 // 求 bucket num 时,将 hash 与 m 相与, // 达到 bucket num 由 hash 的低 8 位决定的效果 m := uintptr(1)<<h.B - 1
B为5时 bucket num 由 hash 的低 5 位决定
// 比如 B=5,那 m 就是31,二进制是全 1 // 求 bucket num 时,将 hash 与 m 相与, // 达到 bucket num 由 hash 的低 5 位决定的效果 m := uintptr(1)<<h.B - 1
The text was updated successfully, but these errors were encountered:
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实际描述
B=5时 bucket num 由 hash 的低 5 位决定
预期描述
B为5时 bucket num 由 hash 的低 5 位决定
The text was updated successfully, but these errors were encountered: