Convert User Input to Integer

How can I read inputs as numbers?

Solution

Since Python 3, input returns a string which you have to explicitly convert to ints, with int, like this

x = int(input("Enter a number: "))
y = int(input("Enter a number: "))

You can accept numbers of any base and convert them directly to base-10 with the int function, like this

>>> data = int(input("Enter a number: "), 8)
Enter a number: 777
>>> data
511
>>> data = int(input("Enter a number: "), 16)
Enter a number: FFFF
>>> data
65535
>>> data = int(input("Enter a number: "), 2)
Enter a number: 10101010101
>>> data
1365

The second parameter tells what is the base of the numbers entered and then internally it understands and converts it. If the entered data is wrong it will throw a ValueError.

>>> data = int(input("Enter a number: "), 2)
Enter a number: 1234
Traceback (most recent call last):
  File "<input>", line 1, in <module>
ValueError: invalid literal for int() with base 2: '1234'

For values that can have a fractional component, the type would be float rather than int:

x = float(input("Enter a number:"))

Differences between Python 2 and 3

Summary

• Python 2's input function evaluated the received data, converting it to an integer implicitly (read the next section to understand the implication), but Python 3's input function does not do that anymore.
• Python 2's equivalent of Python 3's input is the raw_input function.

Python 2.x

There were two functions to get user input, called input and raw_input. The difference between them is, raw_input doesn't evaluate the data and returns as it is, in string form. But, input will evaluate whatever you entered and the result of evaluation will be returned. For example,

>>> import sys
>>> sys.version
'2.7.6 (default, Mar 22 2014, 22:59:56) \n[GCC 4.8.2]'
>>> data = input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
(22, <type 'int'>)

The data 5 + 17 is evaluated and the result is 22. When it evaluates the expression 5 + 17, it detects that you are adding two numbers and so the result will also be of the same int type. So, the type conversion is done for free and 22 is returned as the result of input and stored in data variable. You can think of input as the raw_input composed with an eval call.

>>> data = eval(raw_input("Enter a number: "))
Enter a number: 5 + 17
>>> data, type(data)
(22, <type 'int'>)

Note: you should be careful when you are using input in Python 2.x. I explained why one should be careful when using it, in this answer.

But, raw_input doesn't evaluate the input and returns as it is, as a string.

>>> import sys
>>> sys.version
'2.7.6 (default, Mar 22 2014, 22:59:56) \n[GCC 4.8.2]'
>>> data = raw_input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
('5 + 17', <type 'str'>)

Python 3.x

Python 3.x's input and Python 2.x's raw_input are similar and raw_input is not available in Python 3.x.

>>> import sys
>>> sys.version
'3.4.0 (default, Apr 11 2014, 13:05:11) \n[GCC 4.8.2]'
>>> data = input("Enter a number: ")
Enter a number: 5 + 17
>>> data, type(data)
('5 + 17', <class 'str'>)

Convert input str to int in python

By default, the input type is string. To convert it into integer, just put int before input. E.g.

number = int(input("Please guess what number I'm thinking of. HINT: it's between 1 and 30: "))
print(type(number))

Sample of the output:

Please guess what number I'm thinking of. HINT: it's between 1 and 30: 30
<class 'int'>   # it shows that the input type is integer

ALTERNATIVE

# any input is string
number = input("Please guess what number I'm thinking of. HINT: it's between 1 and 30: ")   
try:                      # if possible, try to convert the input into integer
    number = int(number)
except:                   # if the input couldn't be converted into integer, then do nothing
    pass
print(type(number))       # see the input type after processing

Sample of the output:

Please guess what number I'm thinking of. HINT: it's between 1 and 30: 25    # the input is a number 25
<class 'int'>   # 25 is possible to convert into integer. So, the type is integer

Please guess what number I'm thinking of. HINT: it's between 1 and 30: AAA   # the input is a number AAA
<class 'str'>   # AAA is impossible to convert into integer. So, the type remains string

How do I convert a string from an input into an integer? (Python 3.9)

You need to assign the value, because int() doesn't apply on place.

number = int(number)

Python int() function

convert argv[1] to integer, without trusting the user input

Since you are stating that the user input cannot be trusted, I don't recommend using the function atoi, for two reasons:

1. The behavior is undefined if the converted number is not representable as an int (i.e. if the number is out of range).

2. The function will return 0 when the conversion failed. However, when this happens, you will not know whether the user actually entered 0 or whether you got this value due to conversion failure.

Both of these problems can be solved if you use the functon strtol instead of atoi.

In your question, you state that you want to reject the input if it contains any non-digit characters. One problem with the function strtol is that it will skip any leading whitespace characters (e.g. space and tab characters) before attempting to convert a number. If you instead want to reject the input if it contains any leading whitespace characters, then you will have to compare each character with isdigit, before using the function strtol. Doing this will also solve the problem of invalid characters after the number, so that input such as 123abc will get rejected.

Here is a program which will attempt to convert argv[1] to an integer, while performing full input validation:

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <errno.h>

int main( int argc, char *argv[] )
{
    long num;
    char *p;

    //verify that argv[1] is valid
    if ( argc < 2 )
    {
        printf( "not enough parameters!\n" );
        exit( EXIT_FAILURE );
    }

    //verify that argv[1] consists only of digits
    for ( p = argv[1]; *p != '\0'; p++ )
    {
        if ( !isdigit( (unsigned char)*p ) )
        {
            printf( "first program argument must consist only of digits!\n" );
            exit( EXIT_FAILURE );
        }
    }

    //attempt to convert argv[1] to integer
    errno = 0;
    num = strtol( argv[1], &p, 10 );

    //verify that conversion was successful
    if ( p == argv[1] )
    {
        printf( "unable to convert to integer\n" );
        exit( EXIT_FAILURE );
    }

    //verify that no range error occurred
    if ( errno == ERANGE )
    {
        printf( "input is out of range\n" );
        exit( EXIT_FAILURE );
    }

    //everything is ok, so print the result
    printf( "The result is: %ld\n", num );

    return 0;
}

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