def save_tuple(self, object):
write = self.write
- save = self.save
- memo = self.memo
proto = self.proto
- if proto >= 1:
- n = len(object)
- if n <= 3:
- if not object:
- write(EMPTY_TUPLE)
- return
- if proto >= 2:
- for element in object:
- save(element)
- # Subtle. Same as in the big comment below.
- if id(object) in memo:
- get = self.get(memo[id(object)][0])
- write(POP * n + get)
- else:
- write(_tuplesize2code[n])
- self.memoize(object)
- return
+ n = len(object)
+ if n == 0 and proto:
+ write(EMPTY_TUPLE)
+ return
+
+ save = self.save
+ memo = self.memo
+ if n <= 3 and proto >= 2:
+ for element in object:
+ save(element)
+ # Subtle. Same as in the big comment below.
+ if id(object) in memo:
+ get = self.get(memo[id(object)][0])
+ write(POP * n + get)
+ else:
+ write(_tuplesize2code[n])
+ self.memoize(object)
+ return
# proto 0, or proto 1 and tuple isn't empty, or proto > 1 and tuple
# has more than 3 elements.
for element in object:
save(element)
- if object and id(object) in memo:
+ if n and id(object) in memo:
# Subtle. d was not in memo when we entered save_tuple(), so
# the process of saving the tuple's elements must have saved
# the tuple itself: the tuple is recursive. The proper action
if proto:
write(POP_MARK + get)
else: # proto 0 -- POP_MARK not available
- write(POP * (len(object) + 1) + get)
+ write(POP * (n+1) + get)
return
# No recursion (including the empty-tuple case for protocol 0).