获得CPU频率(每秒周期数) - 3种不同的方式:

Question

我有三个计算来获得当前的cpu频率:(每秒周期数)

_initialCycles  = rdtsc(); //rdtsc function calculates the cpu cycles since init.


first:
------
unsigned int initialMillisec = (_timeVal.tv_sec)*1000+(_timeVal.tv_usec)/1000;
unsigned int cps1 = ((_initialCycles / initialMillisec) * 1000);

second:
-------
double initialMillisec2 = (_timeVal.tv_sec)*1000+(_timeVal.tv_usec)/1000.0;
unsigned int cps2 = (unsigned int)(_initialCycles / initialMillisec2 * 1000.0);

third:
------
unsigned long long initialUsec3 = 
                ((unsigned long long)(_timeVal.tv_sec))*1000000+_timeVal.tv_usec) ;
unsigned int cps3 = (unsigned int)((_initialCycles / initialUsec3)* 1000000.0);

问题是,cps1 cps2 cps3假设具有相同的值,或多或少..但它们没有.我明白了:

cps1 = 2824048000
cps2 = 3824609671
cps3 = 8000000

有谁知道为什么？

Answer 1

整数类型的问题在于舍入.当你除以某个大数字时,你会丢失有效数字.因此,您应该更改乘法和除法之间的操作顺序.首先,你应该乘以1000(或者在100的情况下long long)然后除以.

使用整数丢失精度的一些示例:

321/20*20 = 320;

321/75*75 = 300;

321/112*112 = 224