@Shady_Shiroe to Programmer [email protected] • 2 years agoCall the Priestimagemessage-square26arrow-up1123arrow-down17
arrow-up1116arrow-down1imageCall the Priest@Shady_Shiroe to Programmer [email protected] • 2 years agomessage-square26
minus-square@nerolink4•2 years agoYou basically got it yeah. You can do x % 2 and check if that is 0, as % calculates the remainder (in python at least)
minus-square@Guy_Fieris_Hairlink2•edit-22 years agoy = 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.
minus-square@[email protected]linkfedilink1•2 years agoint val = 3; bool is_odd = val & (decltype(val))0x01 == (decltype (val))1;
minus-square@nerolinkEnglish1•2 years agoKnow you probably don’t care, but you can do a while loop using while True: <code> 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.
You basically got it yeah. You can do x % 2 and check if that is 0, as % calculates the remainder (in python at least)
y = 1
while y ==1:
x = input(“Enter Number”)
print(“EVEN”)
print(“ODD”)
I don’t know why, but I opened up my IDLE for the first time in a year and did it.
int val = 3; bool is_odd = val & (decltype(val))0x01 == (decltype (val))1;Know you probably don’t care, but you can do a while loop using while True: <code>
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.
Unless you want it to keep asking the question of course**