Functions without a return statement always type-check

[Kerl13]

The result of this function has type NoneType and not int

def f(x):
    """ int -> int """

assert f(0)==f(0)
assert f(1)==f(1)

=== Interprétation de : 'test.py' ===
==> Toutes les fonctions sont testées (bien)
==> le programme est bien typé (très bien)
==> Tous les 2 tests sont passés avec succès

(Mr Python version 06786b5)

[fredokun]

It's not that simple ... because I cannot be sure that the last instruction will be a "return None" because it's potential dead code ... I think I can still give a type error for non-NoneType functions when I do not encounter any return (but I cannot be 100% precise without a control-flow analysis...).

[fredokun]

A solution that is "acceptable", I think, has been push in commit c72b9e1
What do you think of this ?

[Kerl13]

This is better but there are still some unhandled cases

def f(x):
    """ int -> int """
    if x > 5:
        return x

Moreover, I guess you forgot to remove some debugging print in the code:

==> Toutes les fonctions sont testées (bien)
==> le programme est bien typé (très bien)
Call(func=Name(id='f', ctx=Load()), args=[
Num(n=6),
], keywords=[])
Call(func=Name(id='f', ctx=Load()), args=[
Num(n=7),
], keywords=[])
==> Tous les 2 tests sont passés avec succès

[OlivierNicole]

I concur. In this (nonsensical) code from a student, I noticed that when there is no unconditional return at the end, the ill-typed assignment res = L[i:j] is not reported:

def eval_chiffres(L):
    """list[int]->int"""

    if len(L)==0:
        return 0

    else:
        #res:int
        res=0
        #x:int
        x=L[0]
        #i:int
        for i in L:
            #j:int
            for j in L:
                if i>x:
                    i=L[0]
                res=L[i:j]
        return res