The following Python code will result in n (14) being printed, as the for loop is completed.
for n in range(15):
if n == 100:
break
else:
print(n)
However, I want the opposite of this. Is there a way to do a for ... else (or while ... else) loop, but only execute the else code if the loop did break?
There is no explicit for...elseifbreak-like construct in Python (or in any language that I know of) because you can simply do this:
for n in range(15):
if n == 100:
print(n)
break
If you have multiple breaks, put print(n) in a function so you Don't Repeat Yourself.
break statement, then just ... run the code before the break statement
goto
A bit more generic solution using exceptions in case you break in multiple points in the loop and don't want to duplicate code:
try:
for n in range(15):
if n == 10:
n = 1200
raise StopIteration()
if n > 4:
n = 1400
raise StopIteration()
except StopIteration:
print n
I didn't really like the answers posted so far, as they all require the body of the loop to be changed, which might be annoying/risky if the body is really complicated, so here is a way to do it using a flag. Replace _break with found or something else meaningful for your use case.
_break = True
for n in range(15):
if n == 100:
break
else:
_break = False
if _break:
print(n)
Another possibility, if it is a function that does nothing if the loop doesn't find a match, is to return in the else: block:
for n in range(15):
if n == 100:
break
else:
return
print(n)
Use:
for n in range(15):
if n == 100:
break
else:
print("loop successful")
if n != range(15)[-1]:
print("loop failed")
break happens in the last loop iteration. It will print both "loop successful" and "loop failed" in that case.
