92 lines
2.6 KiB
Plaintext
92 lines
2.6 KiB
Plaintext
This rule raises an issue when a loop with an ``++else++`` clause doesn't contain any ``++break++`` statement in its body.
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== Why is this an issue?
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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.
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== How to fix it
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Add a ``++break++`` statement to the loop body containing an ``++else++`` clause or remove the ``++else++`` clause.
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=== Code examples
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==== Noncompliant code example
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[source,python,diff-id=1,diff-type=noncompliant]
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----
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from typing import List
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def foo(elements: List[str]):
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for elt in elements:
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if elt.isnumeric():
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return elt
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else: # Noncompliant: no break in the loop
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raise ValueError("List does not contain any number")
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def bar(elements: List[str]):
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for elt in elements:
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if elt.isnumeric():
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return elt
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else: # Noncompliant: no break in the loop
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raise ValueError("List does not contain any number")
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----
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==== Compliant solution
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[source,python,diff-id=1,diff-type=compliant]
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----
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from typing import List
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def foo(elements: List[str]):
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for elt in elements:
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if elt.isnumeric():
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break
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else:
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raise ValueError("List does not contain any number")
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return elt
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def bar(elements: List[str]):
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for elt in elements:
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if elt.isnumeric():
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return elt
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raise ValueError("List does not contain any number")
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----
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== Resources
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=== Documentation
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* 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]
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ifdef::env-github,rspecator-view[]
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'''
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== Implementation Specification
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(visible only on this page)
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=== Message
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Add a "break" statement or remove this "else" clause.
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'''
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== Comments And Links
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(visible only on this page)
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=== on 14 Apr 2020, 09:49:58 Pierre-Yves Nicolas wrote:
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I would expect the compliant solution to suggest something like:
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----
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for i in range(50):
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if i == 42:
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print('Magic number in range')
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print('Magic number not found')
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----
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That behaves the same way as the noncompliant example but it doesn't contain the suspicious `else`.
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endif::env-github,rspecator-view[]
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