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Overview
Comment:Fix a typo in the file format documentation.
Timelines: family | ancestors | descendants | both | trunk
Files: files | file ages | folders
SHA1: c1bf8efee714b8039612cfe6530a14559b4c94f4
User & Date: drh 2012-05-12 01:50:48
Context
2012-05-12
20:24
Add a line to the change log to describe an optimization that was added, forgotten, then remembered during the pre-release code inspection. check-in: 70f4da8259 user: drh tags: trunk
01:50
Fix a typo in the file format documentation. check-in: c1bf8efee7 user: drh tags: trunk
2012-05-11
17:01
Update the change history and news. And make some minor fixes to requirements text. check-in: 9778992724 user: drh tags: trunk
Changes
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Changes to pages/fileformat2.in.

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<tcl>hd_fragment walcksm {WAL checksum algorithm}</tcl>
<h3>4.2 Checksum Algorithm</h3>

<p>The checksum is computed by interpreting the input as
an even number of unsigned 32-bit integers: x(0) through x(N).
^The 32-bit integers are big-endian if the
magic number in the first 4 bytes of the WAL header is 0x377f0683 and
the integers are little-endian the magic number is 0x377f0682.
^The checksum values are always stored in the frame header in a
big-endian format regardless of which byte order is used to compute
the checksum.</p>

<p>The checksum algorithm only works for content which is a multiple of
8 bytes in length.  In other words, if the inputs are x(0) through x(N)
then N must be odd.







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<tcl>hd_fragment walcksm {WAL checksum algorithm}</tcl>
<h3>4.2 Checksum Algorithm</h3>

<p>The checksum is computed by interpreting the input as
an even number of unsigned 32-bit integers: x(0) through x(N).
^The 32-bit integers are big-endian if the
magic number in the first 4 bytes of the WAL header is 0x377f0683 and
the integers are little-endian if the magic number is 0x377f0682.
^The checksum values are always stored in the frame header in a
big-endian format regardless of which byte order is used to compute
the checksum.</p>

<p>The checksum algorithm only works for content which is a multiple of
8 bytes in length.  In other words, if the inputs are x(0) through x(N)
then N must be odd.