Python has never used braces to define code blocks, it relies on indentation instead; this is one of the defining features of the language. There's even a little cookie that CPython gives you to show how strongly they feel about this:
>>> from __future__ import braces
SyntaxError: not a chance
When I saw this little snippet posted to a forum (since deleted) I thought it cannot possibly work. But it does!
>>> def hi(): {
print('Hello')
}
>>> hi()
Hello
Why does this code work, when it appears to violate the language syntax?
The braces aren't defining a code block as they would in other languages - they're defining a set. The print function is being evaluated and its return value (None) is being placed in the set. Once the set is created it is immediately discarded since it isn't being assigned to anything.
There are a couple of Python syntax features that are being exploited here. First, Python allows a single-statement code block to come immediately after a :. Second, an expression is allowed to span multiple lines under certain circumstances.
This code wouldn't have worked if the body of the block were more than one line, or if an assignment or statement other than a function call were attempted.
Here's a redoing of the function to make it clearer what's happening:
>>> def hi2(): print(
{ print('Hello') }
)
>>> hi2()
Hello
{None}
Related
I'm learning Python and, so far, I absolutely love it. Everything about it.
I just have one question about a seeming inconsistency in function returns, and I'm interested in learning the logic behind the rule.
If I'm returning a literal or variable in a function return, no parentheses are needed:
def fun_with_functions(a, b):
total = a + b
return total
However, when I'm returning the result of another function call, the function is wrapped around a set of parentheses. To wit:
def lets_have_fun():
return(fun_with_functions(42, 9000))
This is, at least, the way I've been taught, using the A Smarter Way to Learn Python book. I came across this discrepancy and it was given without an explanation. You can see the online exercise here (skip to Exercize 10).
Can someone explain to me why this is necessary? Is it even necessary in the first place? And are there other similar variations in parenthetical syntax that I should be aware of?
Edit: I've rephrased the title of my question to reflect the responses. Returning a result within parentheses is not mandatory, as I originally thought, but it is generally considered best practice, as I have now learned.
It's not necessary. The parentheses are used for several reason, one reason it's for code style:
example = some_really_long_function_name() or another_really_function_name()
so you can use:
example = (some_really_long_function_name()
or
another_really_function_name())
Another use it's like in maths, to force evaluation precede. So you want to ensure the excute between parenthese before. I imagine that the functions return the result of another one, it's just best practice to ensure the execution of the first one but it's no necessary.
I don't think it is mandatory. Tried in both python2 and python3, and a without function defined without parentheses in lets_have_fun() return clause works just fine. So as jack6e says, it's just a preference.
if you
return ("something,) # , is important, the ( ) are optional, thx #roganjosh
you are returning a tuple.
If you are returning
return someFunction(4,9)
or
return (someFunction(4,9))
makes no difference. To test, use:
def f(i,g):
return i * g
def r():
return f(4,6)
def q():
return (f(4,6))
print (type(r()))
print (type(q()))
Output:
<type 'int'>
<type 'int'>
Let's say, I've got a function like this:
def myFunc():
# useful function to calculate stuff
This will produce an indentation error, unless I add pass:
def myFunc():
# useful function to calculate stuff
pass
However, if I replace a comment with docstring, no pass is necessary:
def myFunc():
"""useful function to calculate stuff"""
This seems like an odd feature as neither of these are used in the program, as far as I know. So, why does it behave like this?
A comment is outright ignored by the interpreter, so omitting a block after an indent is a syntax error. However, a docstring is a real Python object--at its most basic, a literal str. A lone expression is a valid block of code:
'This is a string. It is a valid (though pretty useless) line of Python code.'
In the case of docstrings in particular, there's also some additional functionality going on, such as being used to set the __doc__ attribute.
>>> def myFunc():
... '''MyDocString'''
...
>>> print(myFunc.__doc__)
MyDocString
Note that this also works for classes:
>>> class MyClass(object):
... '''MyClassDocString'''
...
>>> print(MyClass.__doc__)
MyClassDocString
A docstring isn't just a comment. It actually has meaning to the interpreter. In the case with a docstring, you could do myFunc.__doc__ and actually get your docstring back (In the other case with a pass, the result myFunc.__doc__ would be None).
In other words, you are actually adding some code to the function body to modify it's behavior (in some circumstances), so no pass is necessary.
Is there a way to avoid "expected indent block" errors in Python without adding pass to a function?
def exclamation(s):
# s += "!!!"
# return s
print exclamation("The indented horror")
The above code results in an error on line 5. Yes, remove the comments and the code works fine. However, in debugging stuff I often find myself in a similar situation. Is this just a hang-up of the off-side rule and something I need to get used to or are there ways around this?
There has to be something within your function definition to avoid a SyntaxError.
The issue is that the interpreter will effectively ignore comments during parsing, and so while to a human it might look like something is there, to the parser it is empty.
As jonrsharpe has pointed out in a comment, you can use docstrings to "comment out" your code and have it still work. This is because the docstring is, in effect, a normal string. As such this will be parsed and won't cause a SyntaxError. To take your example code it would look like:
def exclamation(s):
'''s += "!!!"
return s'''
# This should print None as nothing is now returned from the func
print(exclamation("The indented horror"))
I have read the following posts but I am still unsure of something.
Python Compilation/Interpretation Process
Why python compile the source to bytecode before interpreting?
If I have a single Python file myfunctions.py containing the following code.
x = 3
def f():
print x
x = 2
Then, saying $ python myfunctions.py runs perfectly fine.
But now make one small change to the above file. The new file looks as shown below.
x = 3
def f():
print x
x = 2
f() # there is a function call now
This time, the code gives out an error. Now, I am trying to understand this behavior. And so far, these are my conclusions.
Python creates bytecode for x=3
It creates a function object f, quickly scans and has bytecode which talks about the local variables within f's scope but note that the bytecode for all statements in Python are unlikely to have been constructed.
Now, Python encounters a function call, it knows this function call is legitimate because the bare minimum bytecode talking about the function object f and its local variables is present.
Now the interpreter takes the charge of executing the bytecode but from the initial footprint it knows x is a local variable here and says - "Why are you printing before you assign?"
Can someone please comment on this? Thanks in advance. And sorry if this has been addressed before.
When the interpreter reads a function, for each "name" (variable) it encounters, the interpreter decides if that name is local or non-local. The criteria that is uses is pretty simple ... Is there an assignment statement anywhere in the body to that name (barring global statements)? e.g.:
def foo():
x = 3 # interpreter will tag `x` as a local variable since we assign to it here.
If there is an assignment statement to that name, then the name is tagged as "local", otherwise, it gets tagged as non-local.
Now, in your case, you try to print a variable which was tagged as local, but you do so before you've actually reached the critical assignment statement. Python looks for a local name, but doesn't find it so it raises the UnboundLocalError.
Python is very dynamic and allows you to do lots of crazy things which is part of what makes it so powerful. The downside of this is that it becomes very difficult to check for these errors unless you actually run the function -- In fact, python has made the decision to not check anything other than syntax until the function is run. This explains why you never see the exception until you actually call your function.
If you want python to tag the variable as global, you can do so with an explicit global1 statement:
x = 3
def foo():
global x
print x
x = 2
foo() # prints 3
print x # prints 2
1python3.x takes this concept even further an introduces the nonlocal keyword
mgilson got half of the answer.
The other half is that Python doesn't go looking for errors beyond syntax errors in functions (or function objects) it is not about to execute. So in the first case, since f() doesn't get called, the order-of-operations error isn't checked for.
In this respect, it is not like C and C++, which require everything to be fully declared up front. It's kind of like C++ templates, where errors in template code might not be found until the code is actually instantiated.
I was working with generator functions and private functions of a class. I am wondering
Why when yielding (which in my one case was by accident) in __someFunc that this function just appears not to be called from within __someGenerator. Also what is the terminology I want to use when referring to these aspects of the language?
Can the python interpreter warn of such instances?
Below is an example snippet of my scenario.
class someClass():
def __init__(self):
pass
#Copy and paste mistake where yield ended up in a regular function
def __someFunc(self):
print "hello"
#yield True #if yielding in this function it isn't called
def __someGenerator (self):
for i in range(0, 10):
self.__someFunc()
yield True
yield False
def someMethod(self):
func = self.__someGenerator()
while func.next():
print "next"
sc = someClass()
sc.someMethod()
I got burned on this and spent some time trying to figure out why a function just wasn't getting called. I finally discovered I was yielding in function I didn't want to in.
A "generator" isn't so much a language feature, as a name for functions that "yield." Yielding is pretty much always legal. There's not really any way for Python to know that you didn't "mean" to yield from some function.
This PEP http://www.python.org/dev/peps/pep-0255/ talks about generators, and may help you understand the background better.
I sympathize with your experience, but compilers can't figure out what you "meant for them to do", only what you actually told them to do.
I'll try to answer the first of your questions.
A regular function, when called like this:
val = func()
executes its inside statements until it ends or a return statement is reached. Then the return value of the function is assigned to val.
If a compiler recognizes the function to actually be a generator and not a regular function (it does that by looking for yield statements inside the function -- if there's at least one, it's a generator), the scenario when calling it the same way as above has different consequences. Upon calling func(), no code inside the function is executed, and a special <generator> value is assigned to val. Then, the first time you call val.next(), the actual statements of func are being executed until a yield or return is encountered, upon which the execution of the function stops, value yielded is returned and generator waits for another call to val.next().
That's why, in your example, function __someFunc didn't print "hello" -- its statements were not executed, because you haven't called self.__someFunc().next(), but only self.__someFunc().
Unfortunately, I'm pretty sure there's no built-in warning mechanism for programming errors like yours.
Python doesn't know whether you want to create a generator object for later iteration or call a function. But python isn't your only tool for seeing what's going on with your code. If you're using an editor or IDE that allows customized syntax highlighting, you can tell it to give the yield keyword a different color, or even a bright background, which will help you find your errors more quickly, at least. In vim, for example, you might do:
:syntax keyword Yield yield
:highlight yield ctermbg=yellow guibg=yellow ctermfg=blue guifg=blue
Those are horrendous colors, by the way. I recommend picking something better. Another option, if your editor or IDE won't cooperate, is to set up a custom rule in a code checker like pylint. An example from pylint's source tarball:
from pylint.interfaces import IRawChecker
from pylint.checkers import BaseChecker
class MyRawChecker(BaseChecker):
"""check for line continuations with '\' instead of using triple
quoted string or parenthesis
"""
__implements__ = IRawChecker
name = 'custom_raw'
msgs = {'W9901': ('use \\ for line continuation',
('Used when a \\ is used for a line continuation instead'
' of using triple quoted string or parenthesis.')),
}
options = ()
def process_module(self, stream):
"""process a module
the module's content is accessible via the stream object
"""
for (lineno, line) in enumerate(stream):
if line.rstrip().endswith('\\'):
self.add_message('W9901', line=lineno)
def register(linter):
"""required method to auto register this checker"""
linter.register_checker(MyRawChecker(linter))
The pylint manual is available here: http://www.logilab.org/card/pylint_manual
And vim's syntax documentation is here: http://www.vim.org/htmldoc/syntax.html
Because the return keyword is applicable in both generator functions and regular functions, there's nothing you could possibly check (as #Christopher mentions). The return keyword in a generator indicates that a StopIteration exception should be raised.
If you try to return with a value from within a generator (which doesn't make sense, since return just means "stop iteration"), the compiler will complain at compile-time -- this may catch some copy-and-paste mistakes:
>>> def foo():
... yield 12
... return 15
...
File "<stdin>", line 3
SyntaxError: 'return' with argument inside generator
I personally just advise against copy and paste programming. :-)
From the PEP:
Note that return means "I'm done, and have nothing interesting to
return", for both generator functions and non-generator functions.
We do this.
Generators have names with "generate" or "gen" in their name. It will have a yield statement in the body. Pretty easy to check visually, since no method is much over 20 lines of code.
Other methods don't have "gen" in their name.
Also, we do not every use __ (double underscore) names under any circumstances. 32,000 lines of code. Non __ names.
The "generator vs. non-generator" method function is entirely a design question. What did the programmer "intend" to happen. The compiler can't easily validate your intent, it can only validate what you actually typed.