In this example I call fun3() in fun2(), but if something special happen in fun3() is there a way to directly return to fun1()? without having to check this special thing in fun2().
def fun1(num):
res = fun2(num)
# do something if fun3() return an error...
def fun2(num):
res = fun3(num) # if fun3(num) return a status:error return to fun(1)
# something else...
def fun3(num):
if type(num) is not int:
return {'status': 'error'}
else:
# something else...
This is what exceptions are for
def fun1(num):
try:
res = fun2(num)
except Exception:
# specific exception handling
# will continue executing
# something else...
def fun2(num):
res = fun3(num) # if fun3(num) return a status:error return to fun(1)
# something else...
def fun3(num):
if type(num) is not int:
raise Exception('Status Error')
else:
# continue
You should consider using a specific Exception type, rather than just Exception. Fitting for this problem would be a TypeError.
Update is not needed), so can I create specific execption types called fetchWarning() and fetchInfo() ? just to return data in the try except in fun1().Warning module instead.class for custom warning type with fetchWarning(UserWarning) so now I can raise it and handle it into fun1() with except exceptions.fetchWarning() as w :