Input in integers, users writes a char, how to catch exception?

raskol · 03-30-2008, 10:47 AM

edit: sorry i didnt mention this is PYTHON. and it is intended but the forum formats it to being unintended.

i have (nonsense)program like this:

anbr=input("input another number: ")
if anbr < 10:
print "your input + 5 is"
print anbr+5
elif anbr == 22:
print "wow 22 u say"
else:
print "hahaha"
raw_input()

if i input "a" instead of an integer it crashes.

how do i catch an exception like this? while anbr!= (int) or something?

or in this case how to do it?

ca = input("guess wrong or live now: ")
tc = {1: "u diieee",
2: "lol feel the pain",
3: "muppet"}
print tc.get(ca, "congrats u live fish")

theNbomr · 03-30-2008, 11:38 AM

Use one of the scanf()/fscanf()/sscanf() family of functions, with a format string to scan an integer. It returns the number of successfully scanned arguments. If that number is less than 1, it was not able to convert the input stream as specified.
--- rod.

Dan04 · 03-30-2008, 11:41 AM

For Python, you need to put your code in CODE blocks to preserve indentation. I'm assuming you meant:

Code:

anbr = input("input another number: ")
if anbr < 10:
    print "your input + 5 is"
    print anbr + 5
elif anbr == 22:
    print "wow 22 u say"
else:
    print "hahaha"
raw_input()

You can deal with invalid input like this:

Code:

def input_int(prompt=""):
    """
    Ask the user to enter an integer and return the value entered.
    If the input is invalid, ask again.
    """
    while True:
        try:
            return int(raw_input(prompt))
        except ValueError:
            print "You must enter an integer."

anbr = input_int("input another number: ")

Note that I used int(raw_input(prompt)) instead of input(prompt). This is because the input function is unsafe: What if the user types __import__('os').remove('some_important_file')?

Dan04 · 03-30-2008, 11:43 AM

Quote:

Originally Posted by theNbomr

Use one of the scanf()/fscanf()/sscanf() family of functions, with a format string to scan an integer. It returns the number of successfully scanned arguments. If that number is less than 1, it was not able to convert the input stream as specified.
--- rod.

That would work, except for the fact that the code isn't written in C

angrybanana · 03-30-2008, 12:28 PM

read it using raw_input() and use .isdigit() to check if it's an int before converting it.
example:

Code:

anbr = raw_input("input another number: ")
while not anbr.isdigit():
    anbr = raw_input('"%s" is not a number, input a NUMBER: ' % anbr)
anbr = int(anbr)
if anbr < 10:
    print "your input + 5 is"
    print anbr + 5
elif anbr == 22:
    print "wow 22 u say"
else:
    print "hahaha"
raw_input()

EDIT: Be careful when using eval it can cause problems. input() uses eval.