小编Oli*_*ung的帖子

在阅读了paxos和筏纸之后,我有以下困惑:paxos论文仅描述了单日志条目的共识,这相当于筏算法的领导者选举部分.在木筏领导者选举中,paxos的方法优于简单的随机超时方法有什么优势？

以下 C++ 程序是否包含任何未定义的行为？

int
main()
{
        struct entry
        {
                uint32_t hash;
                uint32_t idx;
        };
        entry arr[31] = {
            {   7978558,  0}, {   9241630,  1}, {  65706826,  2},
            { 639636154,  3}, {1033996244,  4}, {1225598536,  5},
            {1231940272,  6}, {1252372402,  7}, {2019146042,  8},
            {1520971906,  9}, {1532931792, 10}, {1818609302, 11},
            {1971583702, 12}, {2116478830, 13}, { 883396844, 14},
            {1942092984, 15}, {1274626222, 16}, { 333950222, 17},
            {1265547464, 18}, { 965867746, 19}, {1471376532, 20},
            { 398997278, 21}, {1414926784, 22}, {1831587680, 23},
            { 813761492, 24}, { …