How to debug beginners code [closed] - python

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I have 2 questions about my code. Why the program doesn't go in the second if statement. How can I end the loop?
from random import *
SecretNumber=randint(1,5)
Guess=int(input("Please enter Guess: "))
NumberofGuesses=1
SecretNumber=0
while Guess != SecretNumber:
NumberofGuesses=NumberofGuesses+1
if Guess>SecretNumber:
print("Please insert a smaller number")
else:
print("Please insert a bigger number")
if Guess==SecretNumber:
print("Number of Guesses: {0}".format(NumberofGuesses))

Your second if is outside the while loop, so it won't get hit until you guesss the secret number. The loop never ends because you never read another guess.
You also have a problem that you are overriding your random secret number with zero.
You need something like:
import random
SecretNumber=random.randint(1,5)
NumberofGuesses=0
while true:
Guess=int(input("Please enter Guess: "))
NumberofGuesses += 1
if Guess == SecretNumber:
break # Got it!
elif Guess>SecretNumber:
print("Please insert a smaller number")
else:
print("Please insert a bigger number")
print("Number of Guesses: {0}".format(NumberofGuesses))

It's because you're setting SecretNumber to 0. Remove it and it should work.

Related

Why is there a SyntaxError Python [closed]

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I wrote this code snippet:
lowestNumber = int(input("\nWhat would you like your lowest number to be?"))
highestNumber = int(input("What would you like your highest number to be?"))
number = random.randint(lowestNumber, highestNumber)
tries = 0
while tries < 10:
guess = int(input(f'\nEnter a number between', lowestNumber))
if guess == number:
print("You guessed correctly! The number was", number)
break
elif guess < number:
print("Too low!")
elif guess > number:
print("Too high!")
tries += 1
SyntaxError: bad input on line 22 in main.py.
Line 22 was guess = int(input(f'\nEnter a number between', lowestNumber)).
I searched it up on google and got nothing, I pasted it into OpenAI's code fixing and it also didn't help.
How can I fix this error?
When you wrote
guess = int(input(f'\nEnter a number between', lowestNumber))
it passed both the string and lowestNumber into the input function. However, you probably wanted to write something like Enter a number between (lowestNumber) and (highestNumber). To do this, you would have to write
guess = int(input(f'\nEnter a number between {lowestNumber} and {highestNumber}. '))
In my example, it passes in one object, the string, which contains lowestNumber and highestNumber in it. In your example, it passes in two objects, the string and lowestNumber.
The formatting you did in the input functions works in print statements, so the print statements are correct, but the input function is not.

Is there a way to get input from a def function? [closed]

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I was wondering if it's possible to get the only input from a def function.
def choose():
while True:
try:
pick = float(input("Enter any number that isn't 0 "))
if pick != 0:
break
else:
pick = float(input("Try again! Enter any number that isn't 0 "))
except ValueError:
print("Sorry, I didn't understand that.")
continue
else:
break
choose()
I will try to be as clear as possible. Can you take the pick input from choose() and store it somewhere else. Say like when your done inputting the number you want.
Can you run:
print(pick + 15)
or you can't take the input from the choose() at all. I would just like to know. Cause if so I don't even know how to do so. So I would appreciate the advice.
You can't access local variables from outside the function. The function should return the value, and you can assign that to another variable.
def choose():
while True:
try:
pick = float(input("Enter any number that isn't 0 "))
if pick != 0:
return pick
else:
print("Try again. The number has to be non-zero!")
except ValueError:
print("Sorry, I didn't understand that.")
choice = choose()
print(choice + 15)
You also shouldn't ask for input in the else: block, because it's going to ask again when the loop repeats. Just print the error message there, without reading input.
You don't need the continue statement, since loops automatically continue unless the repetition condition becomes false (which can never happen with while True:) or you execute a break or return statement to leave the loop.

How to limit the number of digits input? [closed]

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I am requesting someone to tell me a program that asks a user to input only 9 digits. Number 0 not at start but having 9 digits.
How can I limit the number of digits?
For example:
Num=int(input('please Enter a Number'))
if num?!
Please help me soon.
If i got this right this should work:
a=input()
if len(a)==9 and a[0] != '0' and a.isdigit():
#if you want it to be integer
a=int(a)
do something
Hope it helps.
Well if I understood correctly you need a program that asks the user to enter exactly 9 digits. In that case it's pretty simple.
i = input("Enter a nine digit number: ")
if len(i) != 9:
print("You must enter 9 digits!")
quit()
if not i.isdigit():
print("Your input must be a number!")
quit()
# num is a 9 digit integer
# i is a 9 digit string
num = int(i)
You can do this.
num = input('Please enter a number')
if not num[0] == "0" and len(num) == 9 and num.isdigit():
# You can print a message or cast the num into an integer here
# like this: input_number = int(num)
print(True)
else:
print(False)
So what's happening here is that the program will accept the user input and evaluate if the first digit is NOT 0 AND that the length is exactly 9 characters.

Using random in an if statement [closed]

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Closed 5 years ago.
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I am trying to make a random number guessing game but I cant get the if statement to check if the users input is = to the random number
import random
realNumber = random.randint(1, 50)
print(realNumber)
myNumber = print(input("Guess the number from 1 to 50: "))
if int(myNumber) == realNumber:
print("You win")
else:
print("Nope guess again")
The unintended behavior of your program is due to this line:
myNumber = print(input("Guess the number from 1 to 50: "))
Here, you are trying to assign myNumber to the return value of the print statement (Which is None) and not the value obtained from the input() statement. To fix this, simply remove the print() around the input.
myNumber = input("Guess the number from 1 to 50: ")
Hope this helped!
You don't need the print statement around input.
import random
realNumber = random.randint(1, 50)
print(realNumber)
myNumber = input("Guess the number from 1 to 50: ")
if int(myNumber) == realNumber:
print("You win")
else:
print("Nope guess again")
Note that this code will not work if the user enters something besides an integer, because the int() call will not cast correctly

Keep getting a syntax error on creating a loop [closed]

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num = input(float("Enter a number (enter 'q' to quit): ")
p=0
while num>=1:
p=p+1
Why do I keep getting a syntax error on the first p?
Does this mean I can't assign a numerical value on a variable?
You should close the bracket
num = input(float("Enter a number (enter 'q' to quit): "))
import sys
num = input("Enter a number (enter 'q' to quit)")
try:
num = int(num)
except:
if num == 'q':
sys.exit()
p=0
while (num>=1):
p=p+1
num = num - 1 #otherwise it will go to infinite loop

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