Question

Possible Duplicate:
function making

>>> g=lambda x:x+1
>>> composition=lambda f,g:(lambda x:f(g(x)))
>>> f=g
>>> f=composition(f,g)
>>> f(9)
11
>>> f=composition(f,g)
>>> f=composition(f,g)
>>> f=composition(f,g)
>>> f(9)
14

Instead of using function composition(f,g) ,i did the following

>>> f=g
>>> f=lambda x:f(g(x))
>>> f(8)
RuntimeError: maximum recursion depth exceeded

I just aimed substituting function composition with corresponding lambda expression. I am confused about why f=composition(f,g) is not producing recursion but f=lambda x:f(g(x)) does.

Answer 1

For the first one:

When you call composition(f, g), you are creating a closure around the contents of those variables. The lambda being returned is looking for the variable names f and g in its local scope – they’ll both be references to the g lambda created on the first line.

For the second one:

In the second one, when f is called, it looks for f in its “local” scope, which is actually the global scope – and finds itself, hence creating infinite recursion.

The key here is that the lookup of f happens at execution, not when the lambda is defined.