Guarding arithmetic

[rusi]

On Aug 23, 3:11 pm, Peter Otten <__pete... at web.de> wrote:
> Mark Carter wrote:
> > Suppose I want to define a function "safe", which returns the argument
> > passed if there is no error, and 42 if there is one. So the setup is
> > something like:
>
> > def safe(x):
> >    # WHAT WOULD DEFINE HERE?
>
> > print safe(666) # prints 666
> > print safe(1/0) # prints 42
>
> > I don't see how such a function could be defined. Is it possible?
>
> 1/0 is evaluated before safe() is called. Therefore safe() has no chance to
> catch the exception. You have to move the evaluation into the safe()
> function:
>
> >>> def safe(deferred, default=42, exception=Exception):
>
> ...     try:
> ...             return deferred()
> ...     except exception:
> ...             return default
> ...>>> print safe(lambda: 666)
> 666
> >>> print safe(lambda: 1/0)
>
> 42

Nice!

Functional programmers know that once you have a lazy language, you
can simulate all control constructs within that framework. eg in
Haskell one could define an 'if-function' as
iffunc (True,  a, b) = a
iffunc (False, a, b) = b
In most other languages such a definition would not work.

What Peter has demonstrated is that for an eager language like python,
a bare-lambda is a lazy-fying construct