rspec/rules/S2836/python/rule.adoc

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.


== Noncompliant Code Example

[source,python]
----
from typing import List

def search_first_number_without_break(elements: List[str]):
    for elt in elements:
        if elt.isnumeric():
            return elt
    else:  # Noncompliant. This will be executed every time
        raise ValueError("List does not contain any number")
----


== Compliant Solution

[source,python]
----
from typing import List

def search_first_number_with_break(elements: List[str]):
    for elt in elements:
        if elt.isnumeric():
            break
    else:
        raise ValueError("List does not contain any number")
    return elt
----
or

[source,python]
----
from typing import List

def search_first_number_without_else(elements: List[str]):
    for elt in elements:
        if elt.isnumeric():
            return elt
    raise ValueError("List does not contain any number")
----


== See

* https://docs.python.org/3/tutorial/controlflow.html#break-and-continue-statements-and-else-clauses-on-loops[Python documentation - break and continue Statements, and else Clauses on Loops]



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