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I think this is not a bug. Here is a simpler way to illustrate the 
issue:
def f(x): 
 for i in range(10): 
 if x: 
 pass 
 continue 
f(True)
f(False)
If you run the code above under trace, you get the following coverage:
 1: def f(x):
 22: for i in range(10):
 20: if x:
 10: pass
 10: continue 
 1: f(True)
 1: f(False)
Note that the 'continue' line is executed 10 instead of expected 20 
times. This happens exactly as Amaury explained. If you disassemble f, 
you'll see
 2 0 SETUP_LOOP 34 (to 37)
 3 LOAD_GLOBAL 0 (range)
 6 LOAD_CONST 1 (10)
 9 CALL_FUNCTION 1
 12 GET_ITER 
 >> 13 FOR_ITER 20 (to 36)
 16 STORE_FAST 1 (i)
 3 19 LOAD_FAST 0 (x)
 22 JUMP_IF_FALSE 4 (to 29)
 25 POP_TOP 
 4 26 JUMP_ABSOLUTE 13
 >> 29 POP_TOP 
 5 30 JUMP_ABSOLUTE 13
 33 JUMP_ABSOLUTE 13
 >> 36 POP_BLOCK 
 >> 37 LOAD_CONST 0 (None)
 40 RETURN_VALUE 
Note how peephole optimizer replaced jump to the 'continue' line (5) 
from the 'pass' line (4) with a jump to the 'for' line by replacing
 4 26 JUMP_FORWARD 1 (to 30)
with
 4 26 JUMP_ABSOLUTE 13
I say this is not a bug because trace is correct in showing that the 
continue statement is never reached when executing f(True).