Reading an Integer from User Input

Reading an integer from user input

You can convert the string to integer using Convert.ToInt32() function

int intTemp = Convert.ToInt32(Console.ReadLine());

How to read an integer input from the user input in dart

The readLineSync() method returns a String?. The ? symbol indicates this variable may be null.

On the other hand, the int.parse() method expects a String, without the ? symbol. This means it doesn't know how to handle if the result from the readLine method comes null.

The easiest way to solve this is to give a default value in case the result comes null:

int num1 = int.parse(stdin.readLineSync() ?? '0');

The ?? operator makes the expression evaluates to the right side, if the left side is null. So giving it a default value it won't have to bother with a nullable type.

There are other operators you can use. You can read more about it in the Dart documentation about it.

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'>)

Reading an integer from input and assigning it to a variable

I don't have a Rust compiler on this machine, but based in part on this answer that comes close, you want something like...

let user_val = match input_string.parse::<i32>() {
    Ok(x) => x,
    Err(_) => -1,
};

Or, as pointed out in the comments,

let user_val = input_string.parse::<i32>().unwrap_or(-1);

...though your choice in integer size and default value might obviously be different, and you don't always need that type qualifier (::<i32>) for parse() where the type can be inferred from the assignment.

Reading an integer from standard input

http://golang.org/pkg/fmt/#Scanf

All the included libraries in Go are well documented.

That being said, I believe

func main() {
    var i int
    _, err := fmt.Scanf("%d", &i)
}

does the trick

Reading user input as an integer

what happends when you call this

    ; getting user input
    mov rax, 0
    mov rdi, 0
    mov rsi, number
    mov rdx, 100
    syscall

is, that your entry ( e.g. "1004") is written to the memory at "number", character per character. Now you have the exact opposite problem you intended to solve:
"how to convert an ASCII string into binary value"

the algorithm for this new problem could look like this:

(assuming char_ptr points to the string)
result = 0;
while ( *char_ptr is a digit )
    result *= 10;
    result += *char_ptr - '0' ;
    char_ptr++;

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