How to Convert String to Int

How do I convert a String to an int in Java?

String myString = "1234";
int foo = Integer.parseInt(myString);

If you look at the Java documentation you'll notice the "catch" is that this function can throw a NumberFormatException, which you can handle:

int foo;
try {
   foo = Integer.parseInt(myString);
}
catch (NumberFormatException e) {
   foo = 0;
}

(This treatment defaults a malformed number to 0, but you can do something else if you like.)

Alternatively, you can use an Ints method from the Guava library, which in combination with Java 8's Optional, makes for a powerful and concise way to convert a string into an int:

import com.google.common.primitives.Ints;

int foo = Optional.ofNullable(myString)
 .map(Ints::tryParse)
 .orElse(0)

How to convert String to Int in Kotlin?

You could call toInt() on your String instances:

fun main(args: Array<String>) {
    for (str in args) {
        try {
            val parsedInt = str.toInt()
            println("The parsed int is $parsedInt")
        } catch (nfe: NumberFormatException) {
            // not a valid int
        }
    }
}

Or toIntOrNull() as an alternative:

for (str in args) {
    val parsedInt = str.toIntOrNull()
    if (parsedInt != null) {
        println("The parsed int is $parsedInt")
    } else {
        // not a valid int
    }
}

If you don't care about the invalid values, then you could combine toIntOrNull() with the safe call operator and a scope function, for example:

for (str in args) {
    str.toIntOrNull()?.let {
        println("The parsed int is $it")
    }
}

How to convert string to integer in PowerShell

You can specify the type of a variable before it to force its type. It's called (dynamic) casting (more information is here):

$string = "1654"
$integer = [int]$string

$string + 1
# Outputs 16541

$integer + 1
# Outputs 1655

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As an example, the following snippet adds, to each object in $fileList, an IntVal property with the integer value of the Name property, then sorts $fileList on this new property (the default is ascending), takes the last (highest IntVal) object's IntVal value, increments it and finally creates a folder named after it:

# For testing purposes
#$fileList = @([PSCustomObject]@{ Name = "11" }, [PSCustomObject]@{ Name = "2" }, [PSCustomObject]@{ Name = "1" })
# OR
#$fileList = New-Object -TypeName System.Collections.ArrayList
#$fileList.AddRange(@([PSCustomObject]@{ Name = "11" }, [PSCustomObject]@{ Name = "2" }, [PSCustomObject]@{ Name = "1" })) | Out-Null

$highest = $fileList |
    Select-Object *, @{ n = "IntVal"; e = { [int]($_.Name) } } |
    Sort-Object IntVal |
    Select-Object -Last 1

$newName = $highest.IntVal + 1

New-Item $newName -ItemType Directory

Sort-Object IntVal is not needed so you can remove it if you prefer.

[int]::MaxValue = 2147483647 so you need to use the [long] type beyond this value ([long]::MaxValue = 9223372036854775807).

Short way to convert string to int

my_input = int(my_input)

There is no shorter way than using the int function (as you mention)

How to convert a string to number in TypeScript?

Exactly like in JavaScript, you can use the parseInt or parseFloat functions, or simply use the unary + operator:

var x = "32";
var y: number = +x;

All of the mentioned techniques will have correct typing and will correctly parse simple decimal integer strings like "123", but will behave differently for various other, possibly expected, cases (like "123.45") and corner cases (like null).

Table taken from this answer

How can I convert a std::string to int?

In C++11 there are some nice new convert functions from std::string to a number type.

So instead of

atoi( str.c_str() )

you can use

std::stoi( str )

where str is your number as std::string.

There are version for all flavours of numbers:
long stol(string), float stof(string), double stod(string),...
see http://en.cppreference.com/w/cpp/string/basic_string/stol

Converting string to int and then back to string

It's all about the priority of the actions:

int.Parse (str_Val + 1)

In the row above first the addition happens str_Val + 1 outputing 11,111,111 etc.

Then the parsing occurs changing "11" to 11

Then to string occurs changing 11 to "11"

So change your code to

str_Val  = (int.Parse(str_Val)+1).ToString();

This will first convert the string to int, then add two integers and finally convert the integer to string again.

How to convert a string to an integer in JavaScript?

The simplest way would be to use the native Number function:

var x = Number("1000")

If that doesn't work for you, then there are the parseInt, unary plus, parseFloat with floor, and Math.round methods.

parseInt:

var x = parseInt("1000", 10); // you want to use radix 10
    // so you get a decimal number even with a leading 0 and an old browser ([IE8, Firefox 20, Chrome 22 and older][1])

unary plus
if your string is already in the form of an integer:

var x = +"1000";

if your string is or might be a float and you want an integer:

var x = Math.floor("1000.01"); //floor automatically converts string to number

or, if you're going to be using Math.floor several times:

var floor = Math.floor;
var x = floor("1000.01");

If you're the type who forgets to put the radix in when you call parseInt, you can use parseFloat and round it however you like. Here I use floor.

var floor = Math.floor;
var x = floor(parseFloat("1000.01"));

Interestingly, Math.round (like Math.floor) will do a string to number conversion, so if you want the number rounded (or if you have an integer in the string), this is a great way, maybe my favorite:

var round = Math.round;
var x = round("1000"); //equivalent to round("1000",0)

how should I convert string to integer and sum the list?

The str and int data types are immutable, so functions called on them can never modify their values.

Hence the int() function can't modify your i variable in the for-loop nor is it supposed to.

As a result, the int() function is designed to return a new integer, thus you must assign it somewhere or it will be "lost in the void".

I.e.

a = ['1', '2', '3']
total = 0
for i in a:
    total = total + int(i)

print(total)

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Note that it is great practice to learn these, albeit simple, algorithms for operations on lists, strings etc, but the sum() built-in function is there to be used if your in a rush!

a = ['1', '2', '3']
print(sum([int(i) for i in a]))

N.B. I have also used a list-comprehension here that you may not be familiar with; they are really useful, I suggest learning them.

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