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why 64-bit atomic operation must make sure the
pointer address is aligned to 8-byte boundary?
Comment 1:
The caller of 64-bit atomic operation must make sure the
pointer address is aligned to 8-byte boundary.
http://play.golang.org/p/fRMRT2Jzsh
Quote BUGS section of docs for sync/atomic:
BUGS
On x86-32, the 64-bit functions use instructions unavailable before the
Pentium MMX. On both ARM and x86-32, it is the caller's responsibility to
arrange for 64-bit alignment of 64-bit words accessed atomically. The
first word in a global variable or in an allocated struct or slice can be
relied upon to be 64-bit aligned.
the Go project does not use its bug tracker for general discussion or asking questions about the language. The Github bug tracker is only used for tracking bugs and proposals going through the Proposal Process.
Please see the Questions wiki page; it has a list of good places for asking questions. Thanks!
why 64-bit atomic operation must make sure the
pointer address is aligned to 8-byte boundary?
Comment 1:
Status changed to WorkingAsIntended.
Originally posted by @minux in #5278 (comment)
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