Unnecessary "else" after "return", remove the "else" and de-indent the code inside it
148 """Enable the auto merge"""
149 sub_run.update(title="Enabling auto-merge", status=CheckRunStatus.IN_PROGRESS)
150 default_branch = repository.get_branch(repository.default_branch)
151 if default_branch.protected:152 if pull_request := pull_request_helper.get_existing_pull_request(
153 repository, branch_name
154 ):Unnecessary "else" after "return", remove the "else" and de-indent the code inside it
127) -> bool:
128 """Enable the auto merge if is enabled and is not the default branch"""
129 if Config.pull_request_manager.enable_auto_merge:
130 if branch == repository.default_branch:131 ignoring_title = f"In the default branch '{branch}', ignoring."
132 sub_run.update(
133 title=ignoring_title,Unnecessary "else" after "return", remove the "else" and de-indent the code inside it
67) -> bool:
68 """Try to create a Pull Request if is enabled and is not the default branch"""
69 if Config.pull_request_manager.create_pull_request:
70 if branch == repository.default_branch: 71 ignoring_title = f"In the default branch '{branch}', ignoring."
72 sub_run.update(
73 title=ignoring_title,
Description
The use of else or elif becomes redundant and can be dropped if the last statement under the leading if / elif block is a return statement.
In the case of an elif after return, it can be written as a separate if block.
For else blocks after return, the statements can be shifted out of else. Please refer to the examples below for reference.
Refactoring the code this way can improve code-readability and make it easier to maintain.
Bad practice
def classify_number(x):
if x % 2 == 0:
return 'Even'
else:
return 'Odd'
def what_is_this_number(x):
if x % 2 == 0 and x >= 0:
return 'Even'
elif x % 2 == 0 and x < 0:
return 'Even and Negative'
elif x % 2 != 0 and x < 0:
return 'Odd and Negative.'
else:
return 'Odd'
Preferred:
def classify_number(x):
if x % 2 == 0:
return 'Even'
return 'Odd'
def what_is_this_number(x):
if x % 2 == 0 and x >= 0:
return 'Even'
if x % 2 == 0 and x < 0:
return 'Even and Negative'
if x % 2 != 0 and x < 0:
return 'Odd and Negative'
return 'Odd'