I am following the solutions from here:
How can I return a JavaScript string from a WebAssembly function
and here:
How to return a string (or similar) from Rust in WebAssembly?
However, when reading from memory I am not getting the desired results.
AssemblyScript file, helloWorldModule.ts:
export function getMessageLocation(): string {
return "Hello World";
}
index.html:
<script>
fetch("helloWorldModule.wasm").then(response =>
response.arrayBuffer()
).then(bytes =>
WebAssembly.instantiate(bytes, {imports: {}})
).then(results => {
var linearMemory = results.instance.exports.memory;
var offset = results.instance.exports.getMessageLocation();
var stringBuffer = new Uint8Array(linearMemory.buffer, offset, 11);
let str = '';
for (let i=0; i<stringBuffer.length; i++) {
str += String.fromCharCode(stringBuffer[i]);
}
debugger;
});
</script>
This returns an offset of 32. And finally yields a string that starts too early and has spaces between each letter of "Hello World":
However, if I change the array to an Int16Array, and add 8 to the offset (which was 32), to make an offset of 40. Like so:
<script>
fetch("helloWorldModule.wasm").then(response =>
response.arrayBuffer()
).then(bytes =>
WebAssembly.instantiate(bytes, {imports: {}})
).then(results => {
var linearMemory = results.instance.exports.memory;
var offset = results.instance.exports.getMessageLocation();
var stringBuffer = new Int16Array(linearMemory.buffer, offset+8, 11);
let str = '';
for (let i=0; i<stringBuffer.length; i++) {
str += String.fromCharCode(stringBuffer[i]);
}
debugger;
});
</script>
Then we get the correct result:
Why does the first set of code not work like its supposed to in the links I provided? Why do I need to change it to work with Int16Array to get rid of the space between "H" and "e" for example? Why do I need to add 8 bytes to the offset?
In summary, what on earth is going on here?
Edit: Another clue, is if I use a TextDecoder on the UInt8 array, decoding as UTF-16 looks more correct than decoding as UTF-8:
AssemblyScript uses utf-16: https://github.com/AssemblyScript/assemblyscript/issues/43
Additionally AssemblyScript stores the length of the string in the first 32 or 64 bits.
That's why my code behaves differently. The examples in the links at the top of this post were for C++ and Rust, which do string encoding differently
I have an angular function which receives a number from a GET request and then will encode that number as base 64. I am trying to use the padStart function outlined here, but when I log the string after calling the function, the string is not padded. Any ideas why?
Angular:
$scope.generateCode = function(){
var number = autogenNumber(); //Pull the number
number.then(function (result) { //Then increment up and convert
var working_str = result.toString();
console.log("The pre padded string is", working_str);
working_str.padStart(8, '1234');
console.log("The padded string is now", working_str, typeof working_str);
})
};
A string is immutable. You have to reassign
working_str = working_str.padStart(8, '1234');
I am using the Scanner methods nextInt() and nextLine() for reading input.
It looks like this:
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:
Enter numerical value
3 // This is my input
Enter 1st string // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string // ...and this line is executed and waits for my input
I tested my application and it looks like the problem lies in using input.nextInt(). If I delete it, then both string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.
That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.
You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).
Workaround:
Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
int option = input.nextInt();
input.nextLine(); // Consume newline left-over
String str1 = input.nextLine();
Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
int option = 0;
try {
option = Integer.parseInt(input.nextLine());
} catch (NumberFormatException e) {
e.printStackTrace();
}
String str1 = input.nextLine();
The problem is with the input.nextInt() method; it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine().
Try it like this, instead:
System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();
It's because when you enter a number then press Enter, input.nextInt() consumes only the number, not the "end of line". When input.nextLine() executes, it consumes the "end of line" still in the buffer from the first input.
Instead, use input.nextLine() immediately after input.nextInt()
There seem to be many questions about this issue with java.util.Scanner. I think a more readable/idiomatic solution would be to call scanner.skip("[\r\n]+") to drop any newline characters after calling nextInt().
EDIT: as #PatrickParker noted below, this will cause an infinite loop if user inputs any whitespace after the number. See their answer for a better pattern to use with skip: https://stackoverflow.com/a/42471816/143585
TL;DR
nextLine() is safe to call when (a) it is first reading instruction, (b) previous reading instruction was also nextLine().
If you are not sure that either of above is true you can use scanner.skip("\\R?") before calling scanner.nextLine() since calls like next() nextInt() will leave potential line separator - created by return key which will affect result of nextLine(). The .skip("\\R?") will let us consume this unnecessary line separator.
skip uses regex where
\R represents line separators
? will make \R optional - which will prevent skip method from:
waiting for matching sequence
in case of reaching end of still opened source of data like System.in, input stream from socket, etc.
throwing java.util.NoSuchElementException in case of
terminated/closed source of data,
or when existing data doesn't match what we want to skip
Things you need to know:
text which represents few lines also contains non-printable characters between lines (we call them line separators) like
carriage return (CR - in String literals represented as "\r")
line feed (LF - in String literals represented as "\n")
when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.
What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows \r\n) after it.
So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42\r\n".
Problem
Scanner#nextInt (and other Scanner#nextType methods) doesn't allow Scanner to consume these line separators. It will read them from System.in (how else Scanner would know that there are no more digits from the user which represent age value than facing whitespace?) which will remove them from standard input, but it will also cache those line separators internally. What we need to remember, is that all of the Scanner methods are always scanning starting from the cached text.
Now Scanner#nextLine() simply collects and returns all characters until it finds line separators (or end of stream). But since line separators after reading the number from the console are found immediately in Scanner's cache, it returns empty String, meaning that Scanner was not able to find any character before those line separators (or end of stream).
BTW nextLine also consumes those line separators.
Solution
So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either
consume line separator left by nextInt from Scanners cache by
calling nextLine,
or IMO more readable way would be by calling skip("\\R") or skip("\r\n|\r|\n") to let Scanner skip part matched by line separator (more info about \R: https://stackoverflow.com/a/31060125)
don't use nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.
BTW: Scanner#nextType methods can skip delimiters (by default all whitespaces like tabs, line separators) including those cached by scanner, until they will find next non-delimiter value (token). Thanks to that for input like "42\r\n\r\n321\r\n\r\n\r\nfoobar" code
int num1 = sc.nextInt();
int num2 = sc.nextInt();
String name = sc.next();
will be able to properly assign num1=42 num2=321 name=foobar.
It does that because input.nextInt(); doesn't capture the newline. you could do like the others proposed by adding an input.nextLine(); underneath.
Alternatively you can do it C# style and parse a nextLine to an integer like so:
int number = Integer.parseInt(input.nextLine());
Doing this works just as well, and it saves you a line of code.
Instead of input.nextLine() use input.next(), that should solve the problem.
Modified code:
public static Scanner input = new Scanner(System.in);
public static void main(String[] args)
{
System.out.print("Insert a number: ");
int number = input.nextInt();
System.out.print("Text1: ");
String text1 = input.next();
System.out.print("Text2: ");
String text2 = input.next();
}
If you want to read both strings and ints, a solution is to use two Scanners:
Scanner stringScanner = new Scanner(System.in);
Scanner intScanner = new Scanner(System.in);
intScanner.nextInt();
String s = stringScanner.nextLine(); // unaffected by previous nextInt()
System.out.println(s);
intScanner.close();
stringScanner.close();
In order to avoid the issue, use nextLine(); immediately after nextInt(); as it helps in clearing out the buffer. When you press ENTER the nextInt(); does not capture the new line and hence, skips the Scanner code later.
Scanner scanner = new Scanner(System.in);
int option = scanner.nextInt();
scanner.nextLine(); //clearing the buffer
If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .
Code :
class ScanReader {
/**
* #author Nikunj Khokhar
*/
private byte[] buf = new byte[4 * 1024];
private int index;
private BufferedInputStream in;
private int total;
public ScanReader(InputStream inputStream) {
in = new BufferedInputStream(inputStream);
}
private int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) return -1;
}
return buf[index++];
}
public char scanChar(){
int c=scan();
while (isWhiteSpace(c))c=scan();
return (char)c;
}
public int scanInt() throws IOException {
int integer = 0;
int n = scan();
while (isWhiteSpace(n)) n = scan();
int neg = 1;
if (n == '-') {
neg = -1;
n = scan();
}
while (!isWhiteSpace(n)) {
if (n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = scan();
}
}
return neg * integer;
}
public String scanString() throws IOException {
int c = scan();
while (isWhiteSpace(c)) c = scan();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = scan();
} while (!isWhiteSpace(c));
return res.toString();
}
private boolean isWhiteSpace(int n) {
if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
else return false;
}
public long scanLong() throws IOException {
long integer = 0;
int n = scan();
while (isWhiteSpace(n)) n = scan();
int neg = 1;
if (n == '-') {
neg = -1;
n = scan();
}
while (!isWhiteSpace(n)) {
if (n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = scan();
}
}
return neg * integer;
}
public void scanLong(long[] A) throws IOException {
for (int i = 0; i < A.length; i++) A[i] = scanLong();
}
public void scanInt(int[] A) throws IOException {
for (int i = 0; i < A.length; i++) A[i] = scanInt();
}
public double scanDouble() throws IOException {
int c = scan();
while (isWhiteSpace(c)) c = scan();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = scan();
}
double res = 0;
while (!isWhiteSpace(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, scanInt());
}
res *= 10;
res += c - '0';
c = scan();
}
if (c == '.') {
c = scan();
double m = 1;
while (!isWhiteSpace(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, scanInt());
}
m /= 10;
res += (c - '0') * m;
c = scan();
}
}
return res * sgn;
}
}
Advantages :
Scans Input faster than BufferReader
Reduces Time Complexity
Flushes Buffer for every next input
Methods :
scanChar() - scan single character
scanInt() - scan Integer value
scanLong() - scan Long value
scanString() - scan String value
scanDouble() - scan Double value
scanInt(int[] array) - scans complete Array(Integer)
scanLong(long[] array) - scans complete Array(Long)
Usage :
Copy the Given Code below your java code.
Initialise Object for Given Class
ScanReader sc = new ScanReader(System.in);
3. Import necessary Classes :
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
4. Throw IOException from your main method to handle Exception
5. Use Provided Methods.
6. Enjoy
Example :
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
class Main{
public static void main(String... as) throws IOException{
ScanReader sc = new ScanReader(System.in);
int a=sc.scanInt();
System.out.println(a);
}
}
class ScanReader....
sc.nextLine() is better as compared to parsing the input.
Because performance wise it will be good.
I guess I'm pretty late to the party..
As previously stated, calling input.nextLine() after getting your int value will solve your problem. The reason why your code didn't work was because there was nothing else to store from your input (where you inputted the int) into string1. I'll just shed a little more light to the entire topic.
Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:
int firstNumber = input.nextInt();
int secondNumber = input.nextInt();
If we input the value below (as a single line of input)
54 234
The value of our firstNumber and secondNumber variable become 54 and 234 respectively. The reason why this works this way is because a new line feed (i.e \n) IS NOT automatically generated when the nextInt() method takes in the values. It simply takes the "next int" and moves on. This is the same for the rest of the nextFoo() methods except nextLine().
nextLine() generates a new line feed immediately after taking a value; this is what #RohitJain means by saying the new line feed is "consumed".
Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.
I hope this helps.. Merry coding!
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
scan.nextLine();
double d = scan.nextDouble();
scan.nextLine();
String s = scan.nextLine();
System.out.println("String: " + s);
System.out.println("Double: " + d);
System.out.println("Int: " + i);
}
if I expect a non-empty input
avoids:
– loss of data if the following input is eaten by an unchecked scan.nextLine() as workaround
– loss of data due to only partially read lines because scan.nextLine() was replaced by scan.next() (enter: "yippie ya yeah")
– Exceptions that are thrown when parsing input with Scanner methods (read first, parse afterwards)
public static Function<Scanner,String> scanLine = (scan -> {
String s = scan.nextLine();
return( s.length() == 0 ? scan.nextLine() : s );
});
used in above example:
System.out.println("Enter numerical value");
int option = input.nextInt(); // read numerical value from input
System.out.println("Enter 1st string");
String string1 = scanLine.apply( input ); // read 1st string
System.out.println("Enter 2nd string");
String string2 = scanLine.apply( input ); // read 2nd string
Use 2 scanner objects instead of one
Scanner input = new Scanner(System.in);
System.out.println("Enter numerical value");
int option;
Scanner input2 = new Scanner(System.in);
option = input2.nextInt();
In one of my usecase, I had the scenario of reading a string value preceded by a couple of integer values. I had to use a "for / while loop" to read the values. And none of the above suggestions worked in this case.
Using input.next() instead of input.nextLine() fixed the issue. Hope this might be helpful for those dealing with similar scenario.
As nextXXX() methods don't read newline, except nextLine(). We can skip the newline after reading any non-string value (int in this case) by using scanner.skip() as below:
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(x);
double y = sc.nextDouble();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(y);
char z = sc.next().charAt(0);
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(z);
String hello = sc.nextLine();
System.out.println(hello);
float tt = sc.nextFloat();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(tt);
Use this code it will fix your problem.
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
input.nextLine();
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
To resolve this problem just make a scan.nextLine(), where scan is an instance of the Scanner object. For example, I am using a simple HackerRank Problem for the explanation.
package com.company;
import java.util.Scanner;
public class hackerrank {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
double d = scan.nextDouble();
scan.nextLine(); // This line shall stop the skipping the nextLine()
String s = scan.nextLine();
scan.close();
// Write your code here.
System.out.println("String: " + s);
System.out.println("Double: " + d);
System.out.println("Int: " + i);
}
}
The nextLine() will read enter directly as an empty line without waiting for the text.
Simple solution by adding an extra scanner to consume the empty line:
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
input.nextLine();
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
This is a very basic problem for beginner coders in java. The same problem I also have faced when I started java (Self Taught).
Actually, when we take an input of integer dataType, it reads only integer value and leaves the newLine(\n) character and this line(i.e. leaved new line by integer dynamic input )creates the problem when we try to take a new input.
eg. Like if we take the integer input and then after try to take an String input.
value1=sc.nextInt();
value2=sc.nextLine();
the value2 will auto read the newLine character and will not take the user input.
Solution:
just we need to add one line of code before taking the next user input i.e.
sc.nextLine();
or
value1=sc.nextInt();
sc.nextLine();
value2=sc.nextLine();
Note: don't forget to close the Scanner to prevent memory leak;
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
scan.nextLine();//to Ignore the rest of the line after (integer input)nextInt()
double d=scan.nextDouble();
scan.nextLine();
String s=scan.nextLine();
scan.close();
System.out.println("String: " + s);
System.out.println("Double: " + d);
System.out.println("Int: " + i);
Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.
int i = new Scanner(System.in).nextInt();
The problem is with the input.nextInt() method - it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine(). Hope this should be clear now.
Try it like that:
System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();
Is it possible to convert mongo objectId into string.
The above pictures shows data i received and shown in console.I need id value in string form .but ObjectId is returning as object
In Database id is look like this- 565d3bf4cefddf1748d1fc5e -objectId and i need id exactly like this –
According to the Mongo documentation:
a 4-byte value representing the seconds since the Unix epoch,
a 3-byte machine identifier,
a 2-byte process id, and
a 3-byte counter, starting with a random value.
You can check it out here: https://docs.mongodb.org/manual/reference/object-id/
So in javascript you could do something like this.
var mId = {
Timestamp:1448950573,
Machine:13565407,
Pid:1756,
Increment:8888962
};
function getId(mongoId) {
var result =
pad0(mongoId.Timestamp.toString(16), 8) +
pad0(mongoId.Machine.toString(16), 6) +
pad0(mongoId.Pid.toString(16), 4) +
pad0(mongoId.Increment.toString(16), 6);
return result;
}
function pad0(str, len) {
var zeros = "00000000000000000000000000";
if (str.length < len) {
return zeros.substr(0, len-str.length) + str;
}
return str;
}
console.log(getId(mId))
It produces "565d3b2dcefddf06dc87a282" which was not exactly the id you had, but that might just be a tweak or i was working with different data :D.
EDIT
Added a padding function so that zeros are not truncated.
Hope that helps
EDIT:
I assume you are using c# to connect to and serve documents from the mongo DB. In that case, there is a driver that also supports toString().
Here is an example using the mongo csharp driver:
using MongoDB.Bson;
using MongoDB.Bson.IO;
using MongoDB.Bson.Serialization;
using MongoDB.Driver;
// ...
string outputFileName; // initialize to the output file
IMongoCollection<BsonDocument> collection; // initialize to the collection to read from
using (var streamWriter = new StreamWriter(outputFileName))
{
await collection.Find(new BsonDocument())
.ForEachAsync(async (document) =>
{
using (var stringWriter = new StringWriter())
using (var jsonWriter = new JsonWriter(stringWriter))
{
var context = BsonSerializationContext.CreateRoot(jsonWriter);
collection.DocumentSerializer.Serialize(context, document);
var line = stringWriter.ToString();
await streamWriter.WriteLineAsync(line);
}
});
}
ORIGINAL:
These are Mongo ObjectId's and if you haven't already deserialised the document they should support a toString method that will return a hexadecimal string.
but if you want this applied to the whole document, using JSON.stringify(MogoDocument) should deserialize this for you into a plain object.
Im trying to control the arduino via using nodejs. My problem is that im trying to write an interger to the arduino but the value won't register. Can any1 help ?
The node.js serial communication code:
var serialport = require("serialport");
var SerialPort = serialport.SerialPort;
var serialPort = new SerialPort("/dev/cu.usbmodem14131", {
baudrate: 9600,
parser: serialport.parsers.readline("\n")
});
serialPort.on("open", function () {
console.log('open');
serialPort.write("45/r/n") // wrinting offset value to the arduino
serialPort.on('data', function(data) {
console.log(data);
});
});
Here the arduino code"
#include <Wire.h>
int offset = 0;
String inString = "";
void setup()
{
Serial.begin(9600);
delay(100);
}
void loop(){
Serial.println(offset); //printing the offset
while (Serial.available() > 0) {
int inChar = Serial.read();
// convert the incoming byte to a char
// and add it to the string:
inString += (char)inChar;
// if you get a newline, print the string,
// then the string's value:
if (inChar == '\n') {
Serial.print("Offset:");
Serial.println(inString.toInt());
offset = inString.toInt();
// clear the string for new input:
inString = "";
}
}
delay(1000);
}
I'm not really sure if i'm writing the value the wrong way or receiving it wrong, but arduino code work fine if I manually enter the value in the arduino IDE.
Thank you.
I'm not sure about your nodejs code but I think there is an issue with the arduino routine. I did notice your nodejs routine is sending /r/n and the Arduino routine is only looking for /n.
The while loop you have can completely execute in less then one serial character time so it may be adding 1 second delays between characters.
I would structure it to stay in the while loop until a newline is received. Following example has not been compiled, but demostrates the concept. It also filters out the '/r' character.
int inChar;
void loop(){
// clear the string for new input:
inString ="";
inChar=0;
while (inchar!= '\n') {
while (Serial.available() > 0) {
inchar = Serial.read();
// do you need to filter the '/r' for it to work correctly?
if (inchar != '/r') {
// convert the incoming byte to a char
// and add it to the string:
inString += (char)inChar;
}
}
}
// when you receive a newline, print the label and
// the string's value:
Serial.print("Offset:");
Serial.println(inString.toInt());
offset = inString.toInt();
delay(1000);
}