CODEWITHSUNDEEP
c#concatenationbitbit-shiftuint16
Rob
Rob

Reputation: 11788

How to bit-shift and concatenate to get correct result?

I'm currently struggling with modbus tcp and ran into a problem with interpreting the response of a module. The response contains two values that are encoded in the bits of an array of three UInt16 values, where the first 8 bits of r[0] have to be ignored.

Let's say the UInt16 array is called r and the "final" values I want to get are val1 and val2, then I would have to do the following:

6

In the above example the desired output values are val1 (=3) and val2 (=6) for the input values r[0]=768, r[1]=1536 and r[2]=0, all values as UInt16.

I already tried to (logically) bit-rightshift r[0] by 8, but then the upper bits get lost because they are stored in the first 8 bits of r[1]. Do I have to concatenate all r-values first and bit-shift after that? How can I do that? Thanks in advance!

Upvotes: 1

Views: 1942

Answers (1)

Jon Skeet
Jon Skeet

Reputation: 1500805

I already tried to (logically) bit-rightshift r[0] by 8, but then the upper bits get lost because they are stored in the first 8 bits of r[1].

Well they're not "lost" - they're just in r[1].

It may be simplest to break it down step by step:

byte val1LowBits = (byte) (r[0] >> 8);
byte val1HighBits = (byte) (r[1] & 0xff);
byte val2LowBits = (byte) (r[1] >> 8);
byte val2HighBits = (byte) (r[2] & 0xff);

uint val1 = (uint) ((val1HighBits << 8) | val1LowBits);
uint val2 = (uint) ((val2HighBits << 8) | val2LowBits);

Upvotes: 2

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