eval() and global variables

[Juan Pablo Romero Méndez]

Hello,

Suppose this function is given:

def f(x,y):
return x+y+k


Is it possible to somehow assign a value to k without resorting to
making k global?

I'm thinking something like this:

eval("f(1,1)", {"f":f, "k":1})

Or even better, something like:

def g(k):
return f

g(1)(1,1) ==> 3


Regards,

Juan Pablo

 

[Peter Otten]

Suppose this function is given:

def f(x,y):
return x+y+k


Is it possible to somehow assign a value to k without resorting to
making k global?

You can replace the function's global dictionary:
.... return x+y+k
....Traceback (most recent call last):
File "<stdin>", line 1, in <module>
NameError: name 'k' is not defined

This is a hack, of course.

Peter

 

[Juan Pablo Romero Méndez]

The hack given by Peter works fine, except in this case:
.... f2 = lambda x,yx,y,fn(x,y))
.... function = type(f2)
.... f3 = function(f2.func_code,dict())
.... print f3
....Traceback (most recent call last):
File "<stdin>", line 1, in <module>


Strange...

 

[Peter Otten]

The hack given by Peter works fine, except in this case:

... f2 = lambda x,yx,y,fn(x,y))
... function = type(f2)
... f3 = function(f2.func_code,dict())
... print f3
...
Traceback (most recent call last):
File "<stdin>", line 1, in <module>



Strange...

The variable fn must be passed to the lambda somehow. You could return f2()
from aaaa(), and f2() would still have to "know" about it. This is called
a "closure".
function(code, globals[, name[, argdefs[, closure]]])

Create a function object from a code object and a dictionary.
The optional name string overrides the name from the code object.
The optional argdefs tuple specifies the default argument values.
The optional closure tuple supplies the bindings for free variables.

In your case we can reuse the closure just like the code:
.... def inner(x, y): return x + fn(x, y)
.... return function(inner.func_code, {}, closure=inner.func_closure)
....8

While we're at it, let's explore the remaining arguments:
210

Peter