I m试图说明如何计算7个轨道和7-13个轨道的2个轴数。
Here is some example c code, just need this in vb.net:
serialBytes[2] = 0x64 & 0x7F; // Second byte holds the lower 7 bits of target.
serialBytes[3] = (0x64 >> 7) & 0x7F; // Third data byte holds the bits 7-13 of target
The 0x7F is a constant so the only number that changes based off input is the 0x64.
Can someone help me out?
该法典将《刑法》转化为《刑法》:
serialBytes(2) = &h64 And &h7F
serialBytes(3) = (&h64 >> 7) And &h7F
如果海产值<代码>64<>>/代码>实际上是一种可变的投入,则仅以输入变量取代<代码>&h64。 如果投入是累进的,那么,你们就必须将结果按顺序排列,但:
serialBytes(2) = CType(value And &H7F, Byte)
serialBytes(3) = CType((value >> 7) And &H7F, Byte)
VB. 该网络没有班轮操作员,但确实有双向操作员。
set bits1to7 = value And &H007F
set bits8to14 = value And &H3F80
<><>Edit>:
VB. 该网络确实有自“网络框架1.1”以来的轨道转移操作者(我的坏人),因此,确实有可能采取更系统的做法:
set bits1to7 = value And &H7F
set bits8to14 = (value >> 7) And &H7F
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