exception - return inside an except python -

what problem when including function inside except? in case have following function:

def inventedfunction(list1):     print "initial list %r" %list1      creates list2 based on list1      try:          list2[1]         print "inside try %r" %list2         inventedfunction(list2)     except:        print "inside except %r" %list2          return list2 

after running inventedfunction(somelist) seems working:

initial list [3, 562, 7, 2, 7, 2, 3, 62, 6] inside try [[3, 562], [2, 7], [2, 7], [3, 62], [6]] initial list [[3, 562], [2, 7], [2, 7], [3, 62], [6]] inside try [[2, 3, 7, 562], [2, 3, 7, 62], [6]] initial list [[2, 3, 7, 562], [2, 3, 7, 62], [6]] inside try [[2, 2, 3, 3, 7, 7, 62, 562], [6]] initial list [[2, 2, 3, 3, 7, 7, 62, 562], [6]] inside except [[2, 2, 3, 3, 6, 7, 7, 62, 562]] 

but not returning anything. if include return list2 outside except returns [[3, 562], [2, 7], [2, 7], [3, 62], [6]] not [[2, 2, 3, 3, 6, 7, 7, 62, 562]] want.

also if change code this:

if len(list2)!=1:     inventedfunction(list2)  else:     return list2 

there no return in function.

another simple example doesn't return anything:

def inventedfunction(list1):     list2=list1[:-1]     if len(list2)!=1:         inventedfunction(list2)      else:         return list2 

you recursively calling function - inventendfunction() - never returning result recursive call, hence not return , need return result returned recursive call -

try:      list2[1]     print "inside try %r" %list2     return inventedfunction(list2)  

also, bad have bare except , should think exception can come when calling inventedfunction() , except exceptions.

---

since in comments -

i guess problem doesn't have function understanding how recursion works.

lets take simple example of function a() recursion , similar yours -

>>> def a(i): ...     if == 1: ...             return "hello" ...     else: ...             a(i+1) ... >>> print(a(0)) none 

as can see above simple example returned 0 , why? lets take step step -

main -> a(0) 

1. you call function a(0) , here i 0 , go else part , , call a(1) .

   main -> a(0) -> a(1) 

2. now, in a again, i 1 , go if part, , returns "hello" .

   main -> a(0)         #returned a(0) 

3. now after return not directly return main() called a(0) , no returns whereever function a(1) called, , inside a(0) , return a(0) , , execution continues, since a(0) not return anything, default return value in main , `none.

---

for example again, need add return inventedfunction(list2) , correctly return results recursive calls, otherwise return value of recursed calls thrown away , not returned back. example -

def inventedfunction(list1):     list2=list1[:-1]     if len(list2)!=1:         return inventedfunction(list2)      else:         return list2