How to exit an if clause

Question

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.

Answer 1

This method works for ifs, multiple nested loops, and other constructs that you can't break from easily.

1. Wrap the code in its own function.
2. Instead of break, use return.

Example:

def some_function():
    if condition_a:
        # do something and return early
        ...
        return
    ...
    if condition_b:
        # do something else and return early
        ...
        return
    ...
    return

if outer_condition:
    ...
    some_function()
    ...

Answer 2

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.)

Answer 3

while some_condition:
   ...
   if condition_a:
       # do something
       break
   ...
   if condition_b:
       # do something
       break
   # more code here
   break

Answer 4

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. :-)

Answer 5

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

Answer 6

For what was actually asked, my approach is to put those ifs inside a one-looped loop

for _ in range(1):
    if (some_condition):
        # do something applicable to all condition
        ...
        if (condition_a):
            # do something
            # and then exit the outer if block
            break
        # do something applicable to some_condition but not for condition_a 
        ...
        if (condition_b):
            # do something
            # and then exit the outer if block
            break
        # do something applicable to some_condition but neither for condition_a/b
        ...

Using this instead of using combination of elif & and, will help you write DRY(don't repeat yourself) code. If you are using elif & and, you might need to rewrite # do something applicable to all condition or other line multiple times. Which is less clean even if the line is just a function call.

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

Answer 7

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.

Answer 8

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.

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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') 

Answer 13

There are several ways to do it. It depends on how the code is implemented.

If you exit from a function, then use return; no code will be executed after the return keyword line. Then the example code is:

def func1(a):
    if a > 100:
        # some code, what you want to do
        return a*a 
    if a < 100:
         # some code, what you want to do
        return a-50
    if a == 100:
         # some code, what you want to do
        return a+a

If you exit from a loop, use break. No code will be executed after break keyword. Then the example code is, for while and for loops:

a = 1
while (True):
    if (a == 10):
        # some code, what you want to do
        break
    else:
        a=a+1
        print("I am number", a)

for i in range(5):
    if i == 3:
        break
    print(i)

If you exit from the basic conditional, then you can use the exit() command directly. Then code after the exit() command will not be executed.

NB: This type of code is not preferable. You can use a function instead of this. But I just share the code for example.

The example code is:

if '3K' in FILE_NAME:
        print("This is MODIS 3KM file")
        SDS_NAME = "Optical_Depth_Land_And_Ocean"
        
    elif 'L2' in FILE_NAME:
        print("This is MODIS 10KM file")
        SDS_NAME = "AOD_550_Dark_Target_Deep_Blue_Combined" 
        exit()
    else:
        print("It is not valid MODIS file")

Answer 14

I scrolled through, and nobody mentioned this technique. It's straightforward compared to other solutions and if you do not want to use a try-except statement, this may be your best option.

#!/usr/bin/python3
import sys

foo = 56

if (foo != 67):
    print("ERROR: Invalid value for foo")
    sys.exit(1)

Answer 15

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.