🔒 How is it going out of range?

[surreal anvil]

@radiant socket

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[radiant socket]

Also here is the code: alphabet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']

text = input("Type your message:\n").lower()
shift = int(input("Type the shift number:\n"))
x = list(text)
c=len(x)
e=len(alphabet)
j=0
print(x)
n=0

while j<c:
for a in range(0, e-1):
if alphabet[a] in x[j]:
x[j] = alphabet[a+shift]
j+=1
n=0

print(x)

[brisk marten]

you're indexing by a + shift given shift can be as big as you like (since it's an user input) then it will break

[fallow hemlock]

if alphabet[a] is 'z', then a shift of just 1 will do this

you probably want to wrap around, use modulo for that: 123 % 100 == 23

[brisk marten]

or check if a + shift < len(alphabet)
that depends on what you wanna do exactly

[radiant socket]

hi

sry my sound was off i think

[radiant socket]

hmm i see

[radiant socket]

yah i also saw ai doing something like that

but i also inputted just abc

and then shift of 1 then also it was going out of range?

so i was just wondering why it was doing that

could there be an explanation for that?

it also said to use break function

[fallow hemlock]

you go through the entire alphabet in the for loop, incrementing j every time

thus, you increment j without the while loop check being used

[radiant socket]

but isn't j is getting incremented only if if the if statement is correct?

[fallow hemlock]

yes, you increment it 3 times because of the 'abc'
but then you reach 'd' in the alphabet and you check again, trying to access x[j] again, which is 'abc'[3], which is out of range

[radiant socket]

hm I'm still trying to understand

[fallow hemlock]

this could be fixed by using short-circuiting:
if j < len(x) and alphabet[a] in x[j]:

the and operator causes the latter part to not get evaluated if the first operand is false

[radiant socket]

sry if im slow

[fallow hemlock]

you can make the code easier to read, and perhaps easier to debug, if you don't use 1-letter variable names

[radiant socket]

okay , thanks:)

[fallow hemlock]

there's also no real reason to save these as variables py c=len(x) e=len(alphabet)

[radiant socket]

i think im getting it a bit now

yah i got that haha but i was too lazy to change it again

it tries to check the code again when i increment j thats why its going out of range?

[fallow hemlock]

you get 'abc'[3], which is out of range, yes

[radiant socket]

Owww

I think I get it now

Thanks: D

Is it done? I just leave it like this?

[fallow hemlock]

write !close if you're sure there's nothing else

[radiant socket]

Okayyy ty:)

Have a great day!

!close

[surreal anvil]

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