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Call the Priest (lemmy.world)

submitted 1 year ago by Shady_Shiroe to c/[email protected]

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[–] nero 4 points 1 year ago (1 children)

You basically got it yeah. You can do x % 2 and check if that is 0, as % calculates the remainder (in python at least)

[–] Guy_Fieris_Hair 2 points 1 year ago* (last edited 1 year ago) (2 children)

y = 1

while y ==1:

x = input("Enter Number")

if int(x) % 2 == 0:

print("EVEN")

else:

print("ODD")

I don't know why, but I opened up my IDLE for the first time in a year and did it.

[–] nero 1 points 1 year ago (1 children)

Know you probably don’t care, but you can do a while loop using while True:

Which is the same as you using a statement which equals to true.

Thought i don’t think you need it for what you’re doing currently.

[–] nero 1 points 1 year ago

Unless you want it to keep asking the question of course**

[–] [email protected] 1 points 1 year ago

int val = 3;
bool is_odd = val & (decltype(val))0x01 == (decltype (val))1;