同样的函数，同样的参数，不同的进程，结果不一样

Question

#includeunsignedintgetHashNum(constchar*str,unsignedintlen,unsignedintmaxPrime){registerunsignedintsum=0;registerunsignedinth=0;registerunsignedshort*p=(unsignedshort*)str;registerunsignedshort*s=(unsignedshort*)str;printf("str:%s,len:%d,h:%d,maxPrime:%d\n",p,len,h,maxPrime);while(p-s

Answer 1

对str的操作越界了，unsignedshortp，p++实际往后对（char*）str移动了sizeof(unsignedshort)=4的距离，如果str长度为3，第一次循环就会跑出str的界,从而对p取值就说不定了。
看出你的本意是按blocks对str求hash值（block大小是一个short），但是在实际指针操作过程有点问题。简单起见，你可以按单字节计算，将p，s指针类型改成unsignedchar×。
如果想要实现按blocks计算，可以参考murmurhash方法。
···
unsignedintmurmur3_32(constchar*key,unsignedintlen,unsignedintseed)
{
staticconstunsignedintc1=0xcc9e2d51;
staticconstunsignedintc2=0x1b873593;
staticconstunsignedintr1=15;
staticconstunsignedintr2=13;
staticconstunsignedintm=5;
staticconstunsignedintn=0xe6546b64;
unsignedinthash=seed;
constintnblocks=len/4;
constunsignedint*blocks=(constunsignedint*)key;
inti;
for(i=0;iunsignedintk=blocks[i];
k*=c1;
k=(k<>(32-r1));
k*=c2;
hash^=k;
hash=((hash<>(32-r2)))*m+n;
}
constchar*tail=(constchar*)(key+nblocks*4);
unsignedintk1=0;
switch(len&3){
case3:
k1^=tail[2]<<16;
case2:
k1^=tail[1]<<8;
case1:
k1^=tail[0];
k1*=c1;
k1=(k1<>(32-r1));
k1*=c2;
hash^=k1;
}
hash^=len;
hash^=(hash>>16);
hash*=0x85ebca6b;
hash^=(hash>>13);
hash*=0xc2b2ae35;
hash^=(hash>>16);
returnhash;
}