c++ - bit operation AND -

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it leetcode problem. given array of numbers nums, in 2 elements appear once , other elements appear twice. find 2 elements appear once.

for example: given nums = [1, 2, 1, 3, 2, 5], return [3, 5]. code is:

class solution { public: vector<int> singlenumber(vector<int>& nums) { int axorb=0;     for(auto i:nums) axorb=axorb^i;     int differbit=(axorb&(axorb-1))^axorb;     int group3=0, group5=0;     for(auto i:nums)

if(differbit&i!=0) group5=group5^i;

        else group3=group3^i;         return vector<int>{group3,group5};  } };

submission result wrong answer.

input:[0,0,1,2] output:[3,0] expected:[1,2]

but if change highlighted part 

if(differbit&i) group5=group5^i;

it accepted. spent lot of time thinking still have no idea. maybe type conversion happened? thanks

this has operator precedence.
because in c && , || operators added late given low priority wouldn't break legacy programs.

this stack overflow question has answer why:

from forum: http://bytes.com/topic/c/answers/167377-operator-precedence

the && , || operators added later "short-circuiting" behavior. dennis ritchie admits in retrospect precedence of bitwise operators should have been changed when logical operators added. several hundred kilobytes of c source code in existence @ point , installed base of 3 computers, dennis thought big of change in c language...

here a table showing operator precedence. enter image description here
showing != @ higher priority &.

as can see bitwise & lower != on table, code doing following:

if ( differbit & (i!=0) )

instead of assume meant do:

if ( (differbit & i) != 0 )

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