renaming 'references' to functions can give recursive problems

[Satchidanand Haridas]

peter wrote:

>Hello, nice solution:
>but it puzzles me :)
>
>can anyone tell me why
>-----------correct solution----------------
>def fA(input):
>  return input
>
>def newFA(input, f= fA):
>   return f(input)
>
>fA = newFA
>
>is correct and:
>  
>
 >>> def fA(input):
...     print "inside fA"
...     return input
...
 >>> def newFA(input,f=fA):
...     print "inside newFA"
...     return f(input)
...
 >>> fA = newFA
 >>> fA(2)
inside newFA
inside fA
2

while:

>-------------infinite loop-----------------
>
>def fA(input):
>  return input
>
>def newFA(input):
>   return fA(input)
>
>fA = newFA
>
>gives an infinite recursive loop?
>
>  
>

 >>> def fA(input):
...     print "inside fA"
...     return input
...
 >>> def newFA(input):
...     print "inside newFA"
...     return fA(input)
...
 >>> fA = newFA
 >>> fA(2)
inside newFA
inside newFA
inside newFA
inside newFA


What is happening is that when you call fA (inside newFA) in the second 
case, you are calling newFA because fA is pointing to newFA (hope that 
made sense ;-)). So it was recursive. While in the former case you 
called f, which pointed to fA, but not to newFA. Probably the following 
will make it clearer:


 >>> def fA(input):
...     print "inside fA"
...     return input
...
 >>> def newFA(input,f=fA):
...     print "inside newFA"
...     print "f is pointing to: ",f
...     return f(input)
...
 >>> fA = newFA
 >>> fA(2)
inside newFA
f is pointing to:  <function fA at 0x43123374>
inside fA
2
 >>> fA
<function newFA at 0x43194064>
 >>> newFA
<function newFA at 0x43194064>

Thus f and fA do not point to the same function object when you execute 
the statement fa(2). This f is called once and terminates.


thanks,
Satchit