```
def is_prime(x):
for n in range (2,x-1):
if x % n == 0:
return False
else:
return True
```

May I please have some feedback on what I'm doing wrong here?

```
def is_prime(x):
for n in range (2,x-1):
if x % n == 0:
return False
else:
return True
```

May I please have some feedback on what I'm doing wrong here?

6 is_prime

You have `if`

- `else`

inside the `for`

loop, and each of those have a `return`

, so no matter what, you're going to return during the very first run of the `for`

, i.e. for 2, even though you want to test other numbers.

**edit**: a better solution: ONLY worry about returning `False`

inside the loop. If you reach the end of the loop, you'll know it to be `True`

, so you can just return `True`

after the loop!

(Remember that `return`

ends the function, immediately giving the value given by the return, so any code after in the function won't run.)

I just did the exercise now so I solved it with

```
def is_prime(a):
if a < 2:
return False
if a == 2: # Oops, copied an old version
return True
for b in range(2, a):
if a % b == 0:
return False
return True
```

Nicely done! Makes sense too. Out of curiosity, why is an else: statement not necessary before returning the Boolean?

I think that @trigonometrical mentioned it when he(she?) said that the function ends when something is returned. So if the if statement returns something, that's the end of the function, we don't have to worry about an else statement, because after that, it doesn't matter what happens.

In all honestly, it's an issue of preference.

Clap clap. Here's my **working** overcomplication:

```
def is_prime(x):
if x in (0,1) or x<0:
return False
elif x==2:
return True
else:
n=2
check=[]
while n < x:
if x % n == 0:
check.append(1)
elif x % n != 0:
check.append(0)
n+=1
if sum(check)>0:
return False
else:
return True
```

This is my code,straightforward

def is_prime(x):

```
for n in range(2,x-1):
if x%n==0:
return False
if x<2:
return False
else:
return True
```

print is_prime(5)

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