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This rule raises an issue when a loop with an ``++else++`` clause doesn't contain any ``++break++`` statement in its body.
== Why is this an issue?
The ``++else++`` clause of a loop is skipped when a ``++break++`` is executed in this loop. In other words, a loop with an ``++else++`` but no ``++break++`` statement will always execute the ``++else++`` part (unless of course an exception is raised or ``++return++`` is used). If this is what the developer intended, it would be much simpler to have the ``++else++`` statement removed and its body unindented. Thus having a loop with an ``++else++`` and no ``++break++`` is most likely an error.
== How to fix it
Add a ``++break++`` statement to the loop body containing an ``++else++`` clause or remove the ``++else++`` clause.
=== Code examples
==== Noncompliant code example
[source,python,diff-id=1,diff-type=noncompliant]
----
from typing import List
def foo(elements: List[str]):
for elt in elements:
if elt.isnumeric():
return elt
else: # Noncompliant: no break in the loop
raise ValueError("List does not contain any number")
def bar(elements: List[str]):
for elt in elements:
if elt.isnumeric():
return elt
else: # Noncompliant: no break in the loop
raise ValueError("List does not contain any number")
----
==== Compliant solution
[source,python,diff-id=1,diff-type=compliant]
----
from typing import List
def foo(elements: List[str]):
for elt in elements:
if elt.isnumeric():
break
else:
raise ValueError("List does not contain any number")
return elt
def bar(elements: List[str]):
for elt in elements:
if elt.isnumeric():
return elt
raise ValueError("List does not contain any number")
----
== Resources
=== Documentation
* Python documentation - https://docs.python.org/3/tutorial/controlflow.html#break-and-continue-statements-and-else-clauses-on-loops[Break and continue Statements, and else Clauses on Loops]
ifdef::env-github,rspecator-view[]
'''
== Implementation Specification
(visible only on this page)
=== Message
Add a "break" statement or remove this "else" clause.
'''
== Comments And Links
(visible only on this page)
=== on 14 Apr 2020, 09:49:58 Pierre-Yves Nicolas wrote:
I would expect the compliant solution to suggest something like:
----
for i in range(50):
if i == 42:
print('Magic number in range')
print('Magic number not found')
----
That behaves the same way as the noncompliant example but it doesn't contain the suspicious `else`.
endif::env-github,rspecator-view[]