最終更新日：2022/12/24
If synchronization is always performed in least-first order with respect to object tags, then situations can never arise in which one thread has the synchronization lock for x while waiting for y and another has the lock for y while waiting for x. Instead, they will both obtain the locks in the same order, thus avoiding this form of deadlock.
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元となった例文
If
synchronization
is
always
performed
in
least-first
order
with
respect
to
object
tags,
then
situations
can
never
arise
in
which
one
thread
has
the
synchronization
lock
for
x
while
waiting
for
y
and
another
has
the
lock
for
y
while
waiting
for
x.
Instead,
they
will
both
obtain
the
locks
in
the
same
order,
thus
avoiding
this
form
of
deadlock.
