Converting hex values in buffer to integer

Question

Background: I'm using node.js to get the volume setting from a device via serial connection. I need to obtain this data as an integer value.

I have the data in a buffer ('buf'), and am using readInt16BE() to convert to an int, as follows:

console.log( buf )
console.log( buf.readInt16BE(0) )

Which gives me the following output as I adjust the external device:

<Buffer 00 7e>
126
<Buffer 00 7f>
127
<Buffer 01 00>
256
<Buffer 01 01>
257
<Buffer 01 02>
258

Problem: All looks well until we reach 127, then we take a jump to 256. Maybe it's something to do with signed and unsigned integers - I don't know!

Unfortunately I have very limited documentation about the external device, I'm having to reverse engineer it! Is it possible it only sends a 7-bit value? Hopefully there is a way around this?

Regarding a solution - I must also be able to convert back from int to this format!

Question: How can I create a sequential range of integers when 7F seems to be the largest value my device sends, which causes a big jump in my integer scale?

Thanks :)

Does it make sense for your device to return a negative number? It does look like the device is using that bit as a sign, in which case reading the buffer as a 16 bit value is incorrect. — forrestj
– forrestj, Commented Jan 24, 2016 at 12:47

forrestj · Accepted Answer · 2016-01-25 06:39:21Z

1

127 is the maximum value of a signed 8-bit integer. If the integer is overflowing into the next byte at 128 it would be safe to assume you are not being sent a 16 bit value, but rather 2 signed 8-bit values, and reading the value as a 16-bit integer would be incorrect.

I would start by using the first byte as a multiplier of 128 and add the second byte, this will give the series you are seeking.

buf = Buffer([0,127]) //<Buffer 00 7f>
buf.readInt8(0) * 128 + buf.readInt8(1)
>127

buf = Buffer([1,0]) //<Buffer 01 00>
buf.readInt8(0) * 128 + buf.readInt8(1)
>128

buf = Buffer([1,1]) //<Buffer 01 01>
buf.readInt8(0) * 128 + buf.readInt8(1)
>129

The way to get back is to divide by 128, round it down to the nearest integer for the first byte, and the second byte contains the remainder.

i = 129
buf = Buffer([Math.floor(i / 128), i % 128])
<Buffer 01 01>

edited Jan 25, 2016 at 6:39

answered Jan 24, 2016 at 13:01

forrestj

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2 Comments

gef Over a year ago

Thanks so much - This perfectly solved the issue for me. Is there a way to do the same in reverse, so 132 == 0x00 and 0x04 ?

forrestj Over a year ago

I see you got it sorted, I edited my post when I had more time to be more clear. Good to see you got it going :)

gef · Accepted Answer · 2016-01-24 20:54:19Z

0

Needed to treat the data as two signed 8-bit values. As per @forrestj the solution is to do:

valueInt = buf.readInt8(0) * 128 + buf.readInt8(1)

We can also convert the int value into the original format by doing the following:

byte1 = Math.floor(valueInt / 128)
byte2 = valueInt % 128

answered Jan 24, 2016 at 20:54

gef

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