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I am writing some code to import data from a stream file. The numeric data is held in little-endian byte order. Is there an easy way to handle/convert this data to big-endian byte order?
It's simply a matter of reversing the order of the bytes in the integer. There's no need to use multiplication or anything like that, just flip the bytes. For example:
midrange.com
*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
* Flip the bytes in a 32-bit integer. If the original was
* in little-endian format, this converts to big-endian,
* and vice-versa.
*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
P flipInt B
D flipInt PI 10I 0
D peInt 10I 0 value
D ds
D Int 10I 0
D Byte1 1A overlay(Int:1)
D Byte2 1A overlay(Int:2)
D Byte3 1A overlay(Int:3)
D Byte4 1A overlay(Int:4)
D save s 1A
/free
Int = peInt;
Save = Byte4;
Byte4 = Byte1;
Byte1 = Save;
Save = Byte3;
Byte3 = Byte2;
Byte2 = Save;
return Int;
/end-free
P E
That code is from a program I'm using in production, so I know that it
works. I suspect that simply copying these bytes like that is much faster
than the multiplying and exponent code that you're using.
I also wrote a proof of concept program that would do this with any field (character or numeric) and in any size (so it'd work with 2, 4, or 8 byte integers, among other things). This one is not used in production anywhere, but I believe that it works correctly. It's probably not as fast as the "flipInt" example, though.
H DFTACTGRP(*NO)
D Reverse PR
D Input * value
D Output * value
D Size 10I 0 value
D Test1 s 5I 0
D Test2 s 5I 0
D Test3 s 26A
D Test4 s 26A
D Test5 s 10I 0
D Test6 s 10I 0
*
* test1 = 16384 (x'4000')
* test2 will be 64 (x'0040')
*
c eval test1 = 16384
c callp Reverse( %addr(test1)
c : %addr(test2)
c : %size(test1) )
c dsply test2
*
* test3 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
* test4 will be 'ZYXWVUTSRQPONMLKJIHGFEDCBA'
*
c eval test3 = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
c callp Reverse( %addr(test3)
c : %addr(test4)
c : %size(test3) )
c dsply test4
*
* Test5 = 2130772483 (x'7F010203')
* Test6 will be 50463103 (x'0302017F')
*
c eval Test5 = 2130772483
c callp Reverse( %addr(test5)
c : %addr(test6)
c : %size(test5) )
c dsply test6
c eval *inlr = *on
*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
* Reverse(): This routine copies the bytes from a input buffer
* to an output buffer in reverse order.
*
* Input = pointer to start of input buffer
* Output = pointer to start of output buffer
* Size = size of input buffer. (output buffer must
* be at least this large as well)
*+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
P Reverse B
D Reverse PI
D Input * value
D Output * value
D Size 10I 0 value
D x s 10I 0
D InByte s 1A based(input)
D OutByte s 1A based(output)
c eval Output = Output + (Size - 1)
c for x = 1 to Size
c eval OutByte = InByte
c eval Input = Input + 1
c eval Output = Output - 1
c endfor
P E
You don't need to worry about the sign. The sign bit stays with the most
significant byte of the integer. So on a big-endian system, it'd be the
leftmost byte, and on a little-endian system it'd be the rightmost byte.
No special checking has to be done because it gets moved with the rest of
the byte.
Good luck
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