Re: Tab delimiter
- From: "HeyBub" <heybubNOSPAM@xxxxxxxxx>
- Date: Sat, 28 Oct 2006 10:12:14 -0500
"William M. Klein" <wmklein@xxxxxxxxxxxxxxxxx> wrote in message
news:yJF0h.70246$R52.55878@xxxxxxxxxxxxxxxxxxxxxxxxx
I found the following in the current LRM,
"2. The ORGANIZATION IS LINE SEQUENTIAL clause specifies that the logical
organization of the file is line sequential. The file organization is
determined at the creation of the file and cannot be changed later. The
records in a line sequential file are delimited by delimiters. Each record
can contain only printable characters and record delimiters. One record is
counted as one line."
that doesn't say what HAPPENS to "non-printable" characters, just that
they aren't supported. Later it says,
"1. When a READ statement has been executed for a line sequential file,
some record data may be transferred to the internal format. Refer to
"NetCOBOL User's Guide" for the transfer rules of record data."
and that lead me to the final answer (in the User Guide),
"If there are tab characters in an input record, spaces are inserted to
align the following characters to preset tab positions. The tab positions
are 1, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, and
72. If a tab is present at a character position beyond 72, it is read as a
single space."
I have never used it, but there is a reference to "High-Speed File
Processing" that includes a statement,
"4. If a record read from a line sequential file includes a tab, the tab
code is not replaced by a blank."
When in doubt, do not trust the documentation.
So, I tested the above theory and, indeed, Fijutsu COBOL does replace tabs
in input with blanks to force alignment at bytes/columns 8, 12, 16, etc.
The compiler does NOT change other unprintable characters (x'00' x'01'
x'255') to spaces, but does include them as characters for counting purposes
to get to the next tab position.
.
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