10

I have a specific situation in which I would like to do the following (actually it is more involved than this, but I reduced the problem to the essence):

>>> (lambda e: 1)(0) if (lambda e: True)(0) else (lambda e: 2)(0)
True

which is a difficult way of writing:

>>> 1 if True else 2
1

but in reality '1','True' and '2' are additional expressions that get evaluated and which require the variable 'e', which I set to '0' for this simplified code example.

Note the difference in output from both expressions above, although

>>> (lambda e: 1)(0)
1
>>> (lambda e: True)(0)
True
>>> (lambda e: 2)(0)
2

The funny thing is that this is a special case, because if I replace '1' by '3' I get the expected/desired result:

>>> (lambda e: 3)(0) if (lambda e: True)(0) else (lambda e: 2)(0)
3

It's even correct if I replace '1' by '0' (which could also be a special case since 1==True and 0==False)

>>> (lambda e: 0)(0) if (lambda e: True)(0) else (lambda e: 2)(0)
0

Also, if I replace 'True' by 'not False' or 'not not True', it still works:

>>> (lambda e: 1)(0) if (lambda e: not False)(0) else (lambda e: 2)(0)
1
>>> (lambda e: 1)(0) if (lambda e: not not True)(0) else (lambda e: 2)(0)
1

Another alternative formulation uses the usual if..then..else statement and does not produce the error:

>>> if (lambda e: True)(0):
    (lambda e: 1)(0)
else:
    (lambda e: 2)(0)

1

What explains this behavior? How can I solve this behavior in a nice way (avoid to use 'not not True' or something?

Thanks!

PS: the question revealed a bug in Python, see https://bugs.python.org/issue25843 for the issue tracking.

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14
  • (lambda e: 1) is just a function that always returns 1 when called. (lambda e: 3) returns 3. I don't see any oddities or special cases. Can you clarify what the problem is? I know this is a toy example, but there is no reason to do a lambda like this... the lambda is just an expression that can be added directly to the if. Commented Dec 5, 2015 at 3:03
  • I am working through a book on programming languages. I am writing a piece of software that will output Python-code in a string. So basically, the piece of code that I'm showing is put inside " " and assigned to a variable. This is supposed to be recursive, so the values '1', 'True' and '0' could in general be other expressions that get evaluated. In some cases the expression depends on the variable 'e', so hence it is passed as an argument in general. Therefore I use (lambda e: 1)(0) instead of 1, because the output is not known in general. I hope this makes it clearer... Commented Dec 5, 2015 at 3:09
  • 1
    Note the fact that the lambda-expression is immediately evaluated, so (lambda e: 1)(0) and not (lambda e: 1). Commented Dec 5, 2015 at 3:13
  • 1
    Interestingly, (lambda e: 1)(0) if (lambda e: False)(0) else (lambda e: 1)(0) (meaning the else clause runs) returns 1, not True. Commented Dec 5, 2015 at 3:36
  • 1
    Something is very wrong with Python 3's behavior here. - EDIT: and something is wrong with Ideone, because the output in that link doesn't correspond to that version of the code. Commented Dec 11, 2015 at 1:22

1 Answer 1

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12

I think I figured out why the bug is happening, and why your repro is Python 3 specific.

Code objects do equality comparisons by value, rather than by pointer, strangely enough:

static PyObject *
code_richcompare(PyObject *self, PyObject *other, int op)
{
    ...

    co = (PyCodeObject *)self;
    cp = (PyCodeObject *)other;

    eq = PyObject_RichCompareBool(co->co_name, cp->co_name, Py_EQ);
    if (eq <= 0) goto unequal;
    eq = co->co_argcount == cp->co_argcount;
    if (!eq) goto unequal;
    eq = co->co_kwonlyargcount == cp->co_kwonlyargcount;
    if (!eq) goto unequal;
    eq = co->co_nlocals == cp->co_nlocals;
    if (!eq) goto unequal;
    eq = co->co_flags == cp->co_flags;
    if (!eq) goto unequal;
    eq = co->co_firstlineno == cp->co_firstlineno;
    if (!eq) goto unequal;

    ...

In Python 2, lambda e: True does a global name lookup and lambda e: 1 loads a constant 1, so the code objects for these functions don't compare equal. In Python 3, True is a keyword and both lambdas load constants. Since 1 == True, the code objects are sufficiently similar that all the checks in code_richcompare pass, and the code objects compare the same. (One of the checks is for line number, so the bug only appears when the lambdas are on the same line.)

The bytecode compiler calls ADDOP_O(c, LOAD_CONST, (PyObject*)co, consts) to create the LOAD_CONST instruction that loads a lambda's code onto the stack, and ADDOP_O uses a dict to keep track of objects it's added, in an attempt to save space on stuff like duplicate constants. It has some handling to distinguish things like 0.0, 0, and -0.0 that would otherwise compare equal, but it wasn't expected that they'd ever need to handle equal-but-inequivalent code objects. The code objects aren't distinguished properly, and the two lambdas end up sharing a single code object.

By replacing True with 1.0, we can reproduce the bug on Python 2:

>>> f1, f2 = lambda: 1, lambda: 1.0
>>> f2()
1

I don't have Python 3.5, so I can't check whether the bug is still present in that version. I didn't see anything in the bug tracker about the bug, but I could have just missed the report. If the bug is still there and hasn't been reported, it should be reported.

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9 Comments

Image
Still the same in 3.5. Looks like candidate for a bug report.
Image
@user2357112 : How does this explain the behavior that is observed in my question? You say lambda e:1 loads a constant 1, but why would this expression then return True? Thanks.
Image
@tvo: Its code object compares equal to that of lambda e: True, and the compiler ends up reusing lambda e: True's code object for lambda e: 1 when it really shouldn't.
Image
@user2357112: Hi, I added an additional code example to the question using the usual if statement. This does not show the problem. Is that consistent with your explanation?
Image
@user2357112: I just checked: the same conflict is occurring between (lambda e: False) and (lambda e: 0), which supports your answer. Did you report a bug?
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