57 lines
1.7 KiB
Plaintext
57 lines
1.7 KiB
Plaintext
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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== Noncompliant Code Example
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----
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from typing import List
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def search_first_number_without_break(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. This will be executed every time
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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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----
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from typing import List
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def search_first_number_with_break(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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----
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or
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----
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from typing import List
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def search_first_number_without_else(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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== See
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* 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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ifdef::env-github,rspecator-view[]
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'''
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== Comments And Links
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(visible only on this page)
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include::comments-and-links.adoc[]
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endif::env-github,rspecator-view[]
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