On Sep 15, 2006, at 12:09 PM, Viktoras Didziulis wrote:
What is the fastest way (in Rev), for instance, to insert 3 bytes
of binary
data after the 465th byte of data stored in variable myVar using
put, seek
or whatever else is available? Or is the repeat loop the only way
to do
this?..
If you mean that the bytes after the 465th byte are shifted down,
then the method Mark Smith mentioned would work. If you mean that
you want to replace bytes 466, 467 and 468, then you can do this:
put binDat into char 466 to 468 of myVar
(There is a feature request in bugzilla to allow the word 'byte' in
this context.)
How you create binDat depends on the form of the binary data. Look
at the binaryEncode() function. Exactly how you would use that
depends on whether you need 3 byte numbers, a single 24-bit number or
a byte plus a 16-bit number or something else.
You can also use numToChar() to create a byte that you think of as
binary. However, my opinion on this has changed and I now encourage
use of binaryEncode() even for single bytes. With binaryEncode() you
can encode all three values at once.
If you are working with imageData and are replacing pixels, then this
is the way to go. If you are building a new imageData, don't forget
the 0 byte--build all 4 bytes at once and then append to the variable
with 'put...after...'.
Dar
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Computer programming
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