Define a lambda expression that raises an Exception

Question

How can I write a lambda expression that\'s equivalent to:

def x():
    raise Exception()

The following is not allowed:

y =         


        

Answer 1

If all you want is a lambda expression that raises an arbitrary exception, you can accomplish this with an illegal expression. For instance, lambda x: [][0] will attempt to access the first element in an empty list, which will raise an IndexError.

PLEASE NOTE: This is a hack, not a feature. Do not use this in any (non code-golf) code that another human being might see or use.

Answer 2

How about:

lambda x: exec('raise(Exception(x))')

Answer 3

I'd like to give an explanation of the UPDATE 3 of the answer provided by Marcelo Cantos:

type(lambda: 0)(type((lambda: 0).__code__)(
    1,0,1,1,67,b'|\0\202\1\0',(),(),('x',),'','',1,b''),{}
)(Exception())

Explanation

lambda: 0 is an instance of the builtins.function class.
type(lambda: 0) is the builtins.function class.
(lambda: 0).__code__ is a code object.
A code object is an object which holds the compiled bytecode among other things. It is defined here in CPython https://github.com/python/cpython/blob/master/Include/code.h. Its methods are implemented here https://github.com/python/cpython/blob/master/Objects/codeobject.c. We can run the help on the code object:

Help on code object:

class code(object)
 |  code(argcount, kwonlyargcount, nlocals, stacksize, flags, codestring,
 |        constants, names, varnames, filename, name, firstlineno,
 |        lnotab[, freevars[, cellvars]])
 |  
 |  Create a code object.  Not for the faint of heart.

type((lambda: 0).__code__) is the code class.
So when we say

type((lambda: 0).__code__)(
    1,0,1,1,67,b'|\0\202\1\0',(),(),('x',),'','',1,b'')

we are calling the constructor of the code object with the following arguments:

• argcount=1
• kwonlyargcount=0
• nlocals=1
• stacksize=1
• flags=67
• codestring=b'|\0\202\1\0'
• constants=()
• names=()
• varnames=('x',)
• filename=''
• name=''
• firstlineno=1
• lnotab=b''

You can read about what the arguments mean in the definition of the PyCodeObject https://github.com/python/cpython/blob/master/Include/code.h. The value of 67 for the flags argument is for example CO_OPTIMIZED | CO_NEWLOCALS | CO_NOFREE.

The most importand argument is the codestring which contains instruction opcodes. Let's see what they mean.

>>> import dis
>>> dis.dis(b'|\0\202\1\0')
          0 LOAD_FAST                0 (0)
          2 RAISE_VARARGS            1
          4 <0>

The documentation of opcodes can by found here https://docs.python.org/3.8/library/dis.html#python-bytecode-instructions. The first byte is the opcode for LOAD_FAST, the second byte is its argument i.e. 0.

LOAD_FAST(var_num)
    Pushes a reference to the local co_varnames[var_num] onto the stack.

So we push the reference to x onto the stack. The varnames is a list of strings containing only 'x'. We will push the only argument of the function we are defining to the stack.

The next byte is the opcode for RAISE_VARARGS and the next byte is its argument i.e. 1.

RAISE_VARARGS(argc)
    Raises an exception using one of the 3 forms of the raise statement, depending on the value of argc:
        0: raise (re-raise previous exception)
        1: raise TOS (raise exception instance or type at TOS)
        2: raise TOS1 from TOS (raise exception instance or type at TOS1 with __cause__ set to TOS)

The TOS is the top-of-stack. Since we pushed the first argument (x) of our function to the stack and argc is 1 we will raise the x if it is an exception instance or make an instance of x and raise it otherwise.

The last byte i.e. 0 is not used. It is not a valid opcode. It might as well not be there.

Going back to code snippet we are anylyzing:

type(lambda: 0)(type((lambda: 0).__code__)(
    1,0,1,1,67,b'|\0\202\1\0',(),(),('x',),'','',1,b''),{}
)(Exception())

We called the constructor of the code object:

type((lambda: 0).__code__)(
    1,0,1,1,67,b'|\0\202\1\0',(),(),('x',),'','',1,b'')

We pass the code object and an empty dictionary to the constructor of a function object:

type(lambda: 0)(type((lambda: 0).__code__)(
    1,0,1,1,67,b'|\0\202\1\0',(),(),('x',),'','',1,b''),{}
)

Let's call help on a function object to see what the arguments mean.

Help on class function in module builtins:

class function(object)
 |  function(code, globals, name=None, argdefs=None, closure=None)
 |  
 |  Create a function object.
 |  
 |  code
 |    a code object
 |  globals
 |    the globals dictionary
 |  name
 |    a string that overrides the name from the code object
 |  argdefs
 |    a tuple that specifies the default argument values
 |  closure
 |    a tuple that supplies the bindings for free variables

We then call the constructed function passing an Exception instance as an argument. Consequently we called a lambda function which raises an exception. Let's run the snippet and see that it indeed works as intended.

>>> type(lambda: 0)(type((lambda: 0).__code__)(
...     1,0,1,1,67,b'|\0\202\1\0',(),(),('x',),'','',1,b''),{}
... )(Exception())
Traceback (most recent call last):
  File "<stdin>", line 3, in <module>
  File "", line 1, in 
Exception

Improvements

We saw that the last byte of the bytecode is useless. Let's not clutter this complicated expression needlesly. Let's remove that byte. Also if we want to golf a little we could omit the instantiation of Exception and instead pass the Exception class as an argument. Those changes would result in the following code:

type(lambda: 0)(type((lambda: 0).__code__)(
    1,0,1,1,67,b'|\0\202\1',(),(),('x',),'','',1,b''),{}
)(Exception)

When we run it we will get the same result as before. It's just shorter.

Answer 4

Functions created with lambda forms cannot contain statements.

Answer 5

Actually, there is a way, but it's very contrived.

You can create a code object using the compile() built-in function. This allows you to use the raise statement (or any other statement, for that matter), but it raises another challenge: executing the code object. The usual way would be to use the exec statement, but that leads you back to the original problem, namely that you can't execute statements in a lambda (or an eval(), for that matter).

The solution is a hack. Callables like the result of a lambda statement all have an attribute __code__, which can actually be replaced. So, if you create a callable and replace it's __code__ value with the code object from above, you get something that can be evaluated without using statements. Achieving all this, though, results in very obscure code:

map(lambda x, y, z: x.__setattr__(y, z) or x, [lambda: 0], ["__code__"], [compile("raise Exception", "", "single"])[0]()

The above does the following:

• the compile() call creates a code object that raises the exception;

• the lambda: 0 returns a callable that does nothing but return the value 0 -- this is used to execute the above code object later;

• the lambda x, y, z creates a function that calls the __setattr__ method of the first argument with the remaining arguments, AND RETURNS THE FIRST ARGUMENT! This is necessary, because __setattr__ itself returns None;

• the map() call takes the result of lambda: 0, and using the lambda x, y, z replaces it's __code__ object with the result of the compile() call. The result of this map operation is a list with one entry, the one returned by lambda x, y, z, which is why we need this lambda: if we would use __setattr__ right away, we would lose the reference to the lambda: 0 object!

• finally, the first (and only) element of the list returned by the map() call is executed, resulting in the code object being called, ultimately raising the desired exception.

It works (tested in Python 2.6), but it's definitely not pretty.

One last note: if you have access to the types module (which would require to use the import statement before your eval), then you can shorten this code down a bit: using types.FunctionType() you can create a function that will execute the given code object, so you won't need the hack of creating a dummy function with lambda: 0 and replacing the value of its __code__ attribute.

Answer 6

There is more than one way to skin a Python:

y = lambda: (_ for _ in ()).throw(Exception('foobar'))

---

Lambdas accept statements. Since raise ex is a statement, you could write a general purpose raiser:

def raise_(ex):
    raise ex

y = lambda: raise_(Exception('foobar'))

But if your goal is to avoid a def, this obviously doesn't cut it. It does, however allow you to conditionally raise exceptions, e.g.:

y = lambda x: 2*x if x < 10 else raise_(Exception('foobar'))

---

Alternatively you can raise an exception without defining a named function. All you need is a strong stomach (and 2.x for the given code):

type(lambda:0)(type((lambda:0).func_code)(
  1,1,1,67,'|\0\0\202\1\0',(),(),('x',),'','',1,''),{}
)(Exception())

And a python3 strong stomach solution:

type(lambda: 0)(type((lambda: 0).__code__)(
    1,0,1,1,67,b'|\0\202\1\0',(),(),('x',),'','',1,b''),{}
)(Exception())

---

Thanks @WarrenSpencer for pointing out a very simple answer if you don't care which exception is raised: y = lambda: 1/0.