What sorts of methods exist for prematurely exiting an if clause?
There are times when I'm writing code and want to put a break statement inside of an if clause, only to remember that those can only be used for loops.
Lets take the following code as an example:
if some_condition:
...
if condition_a:
# do something
# and then exit the outer if block
...
if condition_b:
# do something
# and then exit the outer if block
# more code here
I can think of one way to do this: assuming the exit cases happen within nested if statements, wrap the remaining code in a big else block. Example:
if some_condition:
...
if condition_a:
# do something
# and then exit the outer if block
else:
...
if condition_b:
# do something
# and then exit the outer if block
else:
# more code here
The problem with this is that more exit locations mean more nesting/indented code.
Alternatively, I could write my code to have the if clauses be as small as possible and not require any exits.
Does anyone know of a good/better way to exit an if clause?
If there are any associated else-if and else clauses, I figure that exiting would skip over them.
This question is related to
python
control-flow
while some_condition:
...
if condition_a:
# do something
break
...
if condition_b:
# do something
break
# more code here
break
You can emulate goto's functionality with exceptions:
try:
# blah, blah ...
# raise MyFunkyException as soon as you want out
except MyFunkyException:
pass
Disclaimer: I only mean to bring to your attention the possibility of doing things this way, while in no way do I endorse it as reasonable under normal circumstances. As I mentioned in a comment on the question, structuring code so as to avoid Byzantine conditionals in the first place is preferable by far. :-)
Effectively what you're describing are goto statements, which are generally panned pretty heavily. Your second example is far easier to understand.
However, cleaner still would be:
if some_condition:
...
if condition_a:
your_function1()
else:
your_function2()
...
def your_function2():
if condition_b:
# do something
# and then exit the outer if block
else:
# more code here
use return in the if condition will returns you out from the function,
so that you can use return to break the the if condition.
from goto import goto, label if some_condition: ... if condition_a: # do something # and then exit the outer if block goto .end ... if condition_b: # do something # and then exit the outer if block goto .end # more code here label .end
(Don't actually use this, please.)
The only thing that would apply this without additional methods is elif as the following example
a = ['yearly', 'monthly', 'quartly', 'semiannual', 'monthly', 'quartly', 'semiannual', 'yearly']
# start the condition
if 'monthly' in b:
print('monthly')
elif 'quartly' in b:
print('quartly')
elif 'semiannual' in b:
print('semiannual')
elif 'yearly' in b:
print('yearly')
else:
print('final')
So here i understand you're trying to break out of the outer if code block
if some_condition:
...
if condition_a:
# do something
# and then exit the outer if block
...
if condition_b:
# do something
# and then exit the outer if block
# more code here
One way out of this is that you can test for for a false condition in the outer if block, which will then implicitly exit out of the code block, you then use an else block to nest the other ifs to do something
if test_for_false:
# Exit the code(which is the outer if code)
else:
if condition_a:
# Do something
if condition_b:
# Do something
Generally speaking, don't. If you are nesting "ifs" and breaking from them, you are doing it wrong.
However, if you must:
if condition_a:
def condition_a_fun():
do_stuff()
if we_wanna_escape:
return
condition_a_fun()
if condition_b:
def condition_b_fun():
do_more_stuff()
if we_wanna_get_out_again:
return
condition_b_fun()
Note, the functions don't HAVE to be declared in the if statement, they can be declared in advance ;) This would be a better choice, since it will avoid needing to refactor out an ugly if/then later on.
There is another way which doesn't rely on defining functions (because sometimes that's less readable for small code snippets), doesn't use an extra outer while loop (which might need special appreciation in the comments to even be understandable on first sight), doesn't use goto (...) and most importantly let's you keep your indentation level for the outer if so you don't have to start nesting stuff.
if some_condition:
...
if condition_a:
# do something
exit_if=True # and then exit the outer if block
if some condition and not exit_if: # if and only if exit_if wasn't set we want to execute the following code
# keep doing something
if condition_b:
# do something
exit_if=True # and then exit the outer if block
if some condition and not exit_if:
# keep doing something
Yes, that also needs a second look for readability, however, if the snippets of code are small this doesn't require to track any while loops that will never repeat and after understanding what the intermediate ifs are for, it's easily readable, all in one place and with the same indentation.
And it should be pretty efficient.
Here's another way to handle this. It uses a single item for loop that enables you to just use continue. It prevents the unnecessary need to have extra functions for no reason. And additionally eliminates potential infinite while loops.
if something:
for _ in [0]:
# Get x
if not x:
continue
# Get y
if not y:
continue
# Get z
if not z:
continue
# Stuff that depends on x, y, and z
For what was actually asked, my approach is to put those ifs inside a one-looped loop
while (True):
if (some_condition):
...
if (condition_a):
# do something
# and then exit the outer if block
break
...
if (condition_b):
# do something
# and then exit the outer if block
break
# more code here
# make sure it is looped once
break
Test it:
conditions = [True,False]
some_condition = True
for condition_a in conditions:
for condition_b in conditions:
print("\n")
print("with condition_a", condition_a)
print("with condition_b", condition_b)
while (True):
if (some_condition):
print("checkpoint 1")
if (condition_a):
# do something
# and then exit the outer if block
print("checkpoint 2")
break
print ("checkpoint 3")
if (condition_b):
# do something
# and then exit the outer if block
print("checkpoint 4")
break
print ("checkpoint 5")
# more code here
# make sure it is looped once
break
may be this?
if some_condition and condition_a:
# do something
elif some_condition and condition_b:
# do something
# and then exit the outer if block
elif some_condition and not condition_b:
# more code here
else:
#blah
if
Source: Stackoverflow.com