why it is not possible to "SyntaxError: can't assign to function call" in Sage?

[Dima Pasechnik]

Consider the following invalid Python code:

>>> def f(x):

...   a=1

...   f(a)=2

... 

  File "<stdin>", line 3

SyntaxError: can't assign to function call

but Sage  allows f to be defined:

sage: def f(x):

....:     a=1

....:     f(a)=2

....:     

sage: type(f)

<type 'function'>

It doesn't make sense to me to allow this to happen.

In this sense, #10747 is only a half-solution (it would catch things like f(1)=2 inside def f(..): body)

Dima

[Keshav Kini]

Well that's because Sage has a preparser. You could as well say that because f(a) = 2 is invalid Python code, so should it be invalid Sage code, regardless of whether it's a function definition or not.

This, on the other hand, is perfectly valid Python:

>>> def f(x):
...     a = 1
...     def f(a):
...         return a
...     return f(x)
...
>>> f(2)
2
>>>

Your invalid Python above is invalid not because of the semantics of redefining f within f, but just the fact that vanilla Python doesn't have the "f(x) = foo" symbolic 'function' definition syntax Sage does.

If there's anything wrong with defining symbolic functions inside other functions it's this, IMO:

sage: a
---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)

/home/fs/src/ipython/<ipython console> in <module>()

NameError: name 'a' is not defined
sage: def f(x):
....:     p(a) = 3
....:     return "foo"
....:
sage: f(2)
'foo'
sage: a
a
sage:

-Keshav

----
Join us in #sagemath on irc.freenode.net !