replace a byte of a long

Bobb Fwed

I was trying to think of an easy way to replace a single byte of a long, but I didn't get much sleep last night, so I am stuck with this solution:

'' switching out byte[noparse][[/noparse] 2 ]
byte := $80
long := (long & $FF_00_FF_FF) | (byte << 16)

There's got to be an easier way that I am just blind to because of lack of beauty sleep.

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April, 2008: when I discovered the answers to all my micro-computational-botherations!

Post Edited (Bobb Fwed) : 8/6/2009 3:59:27 PM GMT

Mike Green

You could say: longValue.byte[noparse][[/noparse] 2 ] := byteValue

Bobb Fwed

Thanks. I knew there was an easier way, I just forgot about the .byte stuff.

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April, 2008: when I discovered the answers to all my micro-computational-botherations!

Bobb Fwed

OK, what about a variable as an array:
can a byte (or word) be extracted/changed from a long?

Or, alternatively, will two declared variables in a method (PUB Main | v1, v2) always be next to each other in hub memory (i.e. I could just store my information to two adjacent variables rather than an array)?

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April, 2008: when I discovered the answers to all my micro-computational-botherations!

Bobb Fwed

Well, I think I figured out a way to use the array:

PUB Main | out, v[noparse][[/noparse] 2 ]

  v[noparse][[/noparse] 0 ] := $12_23_34_45
  v[noparse][[/noparse] 1 ] := $56_67_78_89

  out := v1.word[noparse][[/noparse] 3 ]

out will be $56_67

EDIT: fixed typo and formatting issue

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April, 2008: when I discovered the answers to all my micro-computational-botherations!

Post Edited (Bobb Fwed) : 8/6/2009 7:05:26 PM GMT

localroger

Bobb, the .byte syntax doesn't work with individual array entries but you can do byte[noparse][[/noparse] @v[noparse][[/noparse] 1 ]·+ 1 ]