Python 3 - EOF Error

Question

I get the following error when I try to execute this code,

Traceback (most recent call last):
  File "c:\dev\tmp\_t636427523447579562.py", line 32, in  tentativa = int(input())
EOFError: EOF when reading a line

import random
import sys

controlo = 0
numero = 134+78



while controlo<3:
    tentativa = int(input())

    if tentativa == numero:
        print("Acertou")
        break
    else:
         controlo +=1
         continue

if controlo == 3:
    print("Falhou")
    sys.exit()



controlo = 0

random_1 = random.randrange(1, 100)
random_2 = random.randrange(1, 100)
soma = random_1 + random_2

while controlo<3:
    tentativa_2 = int(input())

    if tentativa_2 == soma:
         print("Acertou")
         break
    else:
            controlo +=1
            continue

if controlo == 3:
    print("Falhou")

Seems to work fine for me. Maybe try unindenting the lines `controlo +=1, continue` at the end of your code — Vladimir Shevyakov, Oct 04 '17 at 21:31

Abhijit Pritam Dutta · Answer 1 · 2017-10-05T06:34:41.630

0

Change of the below line may fix your problem

tentativa = int(input())

read = input()

now you can check here whether 'read' is digit or mix with character add a if condition and check

tentative = int(read)

edited Oct 05 '17 at 06:34

answered Oct 04 '17 at 22:48

Abhijit Pritam Dutta

5,521
2
11
17

1

In Python there is no difference between these two syntaxs. – blueenvelope Oct 04 '17 at 23:08

score 0 · Answer 2 · answered Oct 04 '17 at 22:59

0

I'm guessing you are using Python 2 here not Python 3 as your tag suggests.

In Python 2 input() waits for the user to enter some text and then tries to execute it. If you press 'Enter' input() will give an EOF error. I assume that is what has happened here as your code works if numbers are entered. I suggest using Python 3 or raw_input() in Python 2. See here for more.

answered Oct 04 '17 at 22:59

blueenvelope

699
7
15

I am using Python 3.6 – Xx_DarDoAzuL_xX Oct 05 '17 at 13:36

Python 3 - EOF Error

2 Answers2