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In my program (written in Python 3.4) I have a variable which contains various flags, so for example:

FLAG_ONE = 0b1
FLAG_TWO = 0b10
FLAG_THREE = 0b100
status = FLAG_ONE | FLAG_TWO | FLAG_THREE

Setting another flag can easily be done with

status |= FLAG_FOUR

But what if I explicitly want to clear a flag? I'd do

status &= ~FLAG_THREE

Is this approach safe? As the size of an integer in Python is not defined, what if status and FLAG_THREE differ in size?

(status needs to be a bit field because I need this value for a hardware protocol.)

John Kugelman
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elzell
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2 Answers2

16

You should be safe using that approach, yes.

~ in Python is simply implemented as -(x+1) (cf. the CPython source) and negative numbers are treated as if they have any required number of 1s padding the start. From the Python Wiki:

Of course, Python doesn't use 8-bit numbers. It USED to use however many bits were native to your machine, but since that was non-portable, it has recently switched to using an INFINITE number of bits. Thus the number -5 is treated by bitwise operators as if it were written "...1111111111111111111011".

In other words, with bitwise-and & you're guaranteed that those 1s will pad the length of ~FLAG (a negative integer) to the length of status. For example:

  100000010000 # status
&       ~10000 # ~FLAG

is treated as

  100000010000
& 111111101111

= 100000000000 # new status

This behaviour is described in a comment in the source here.

Alex Riley
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  • Have you tried to find where this behavior is defined in the official Python docs? – jfs Jan 22 '16 at 18:46
  • I had a look through the documentation pages but couldn't find the main point about the behaviour outlined as clearly as it is in the wiki pages. It's very possible that I overlooked something, of course. – Alex Riley Jan 23 '16 at 13:44
12

Clearing a flag works with

status &= ~FLAG_THREE

because Python treats those negated values as negative:

>>> ~1L
-2L
>>> ~1
-2
>>> ~2
-3

Thus the & operator can act appropriately and yield the wanted result irrespective of the length of the operands, so 0b11111111111111111111111111111111111111111111111111111111111 & ~1 works fine although the left hand operand is biffer than the right hand one.

In the other direction (RH longer than LH), it works nevertheless, because having an excess number of 1 bits doesn't matter.

glglgl
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