So I tried the following code to also print out whether x is a prime or not, instead of just none and False. Why does it print only 4 is not a prime and not 2 is a prime?
def is_prime (x):
if x < 2:
return False
else:
for n in range (2, x-1):
if x % n == 0:
print x, "is not a prime"
return False
else:
print x, "is a prime"
return True
print is_prime (2)
print is_prime (4)
return means handing back a value to the function call, and indicating that the function is done/completed.
your function returns in the first iteration of the loop, so the return will break free from the loop (so to speak)
def is_prime (x):
if x < 2:
return False
else:
for n in range (2, x-1):
if x % n == 0:
print x, “is not a prime”
else:
print x, “is a prime”
is_prime (2)
is_prime (4)
But even like this it does not print 2 is a prime. And I am not returning True or False in between. I don’t understand how I can have it print 2 is a prime as well.
for x = 2 the loop will never run, there are no steps to take.
you solve two problems at once figuring out that a number is prime number after the loop ran all its iteration without the if condition being true, thus you get for else
Somewhere in the documentation it says that when,
for...
else...
is used, there should be a break in the for loop. No break, no else.
In a for - else loop, break causes else to be bypassed, so if you are searching for something and do not find it, you can print or return an appropriate notification. (The else clause also executes if the for loop is bypassed.):
for n in [3,7,9,10]: # break is encountered
if n % 2 == 0:
print("{} is even".format(n))
break
else:
print("no even numbers")
# Output:
10 is even
vs
for n in [3,7,9]: # break is not encountered
if n % 2 == 0:
print("{} is even".format(n))
break
else:
print("no even numbers")
# Output:
no even numbers
vs
for n in []: # No break, but for loop never entered
if n % 2 == 0:
print("{} is even".format(n))
else:
print("no even numbers")
# Output:
no even numbers
Of course, if break is omitted, it won’t work like you want:
for n in [3,7,9, 10]: # No break here
if n % 2 == 0:
print("{} is even".format(n))
else:
print("no even numbers")
# Output:
10 is even
no even numbers
def is_prime(x):
if x < 2:
return False
else:
for n in range(2, x-1):
if x % n == 0:
return False
return True
This is the solution provided. What I don’t understand here is that the final return is by the function itself. In this way, the returned value should always be True, regardless of what the else statement returns?
no. Once a return keyword is reached, the function hands back data and is done executing. So when return false is reached, the function ends. It never reaches return True then