计算根号2的快慢问题

[duckyaya]

试编制一个程序，比较在计算根号2时是用sqr（2）方法快还是用2^(0.5)的方法快，进一步比较二者相差多少倍。

请教中...........

[rainstormmaster]

先算出执行10000次sqr（2）所用时间，除以10000

[flc]

关注

[duckyaya]

怎么计算时间？

[duckyaya]

用timer还是？

[boydgmx]

改天有时间了我试试看，呵呵

[flyingscv]

应该是sqr（2）

[duckyaya]

Private Sub Form_Load()

Print Timer

For i = 0 To 100000
a = Sqr(2)
Next i

Print Timer

For i = 0 To 100000
a = 2 ^ (0.5)
Next i

Print Timer

End Sub

[duckyaya]

sqr快！

[飘零风]

sqr快，我觉得这个不用做实验也能肯定。毕竟，sqr()是专用的开平方函数，而^是通用的。

[boydgmx]

测试表明，很显然sqr快

计算10万次
sqr(2)耗时0.06秒
2 ^ (0.5)耗时0.39秒

[northwolves]

Private Type LARGE_INTEGER
lowpart As Long
highpart As Long
End Type

Private Declare Function QueryPerformanceCounter Lib "kernel32" _
(lpPerformanceCount As LARGE_INTEGER) As Long
Private Declare Function QueryPerformanceFrequency Lib "kernel32" _
(lpFrequency As LARGE_INTEGER) As Long
Dim starttime As LARGE_INTEGER, frequency As LARGE_INTEGER, nowtime As LARGE_INTEGER
Dim nowtime1 As LARGE_INTEGER
Dim nowtime2 As LARGE_INTEGER
Private Sub Command1_Click()
MsgBox timediff, 0, "needtime(2^0.5)-needtime(sqr(2))"
End Sub
Function timediff()
QueryPerformanceFrequency frequency
QueryPerformanceCounter starttime
a = Sqr(2)
QueryPerformanceCounter nowtime1
b = 2 ^ 0.5
QueryPerformanceCounter nowtime2
timediff = Format((nowtime2.lowpart - nowtime1.lowpart) / frequency.lowpart, "0.00000000000000") & "s"
End Function