This is my Code
public class Workshop3
{
public static void main (String [] args)
{
System.out.println ("please enter radius of circle");
double radius;
Scanner keyboard = new Scanner (System.in);
keyboard.nextDouble (radius);
}
}
The error I recieve is
cannot find symbol — class scanner
on the line
Scanner keyboard = new Scanner (System.in);
gunr2171
15.9k25 gold badges61 silver badges87 bronze badges
asked May 11, 2011 at 3:56
0
As the OP is a new beginner to programming, I would like to explain more.
You wil need this line on the top of your code in order to compile:
import java.util.Scanner;
This kind of import statement is very important. They tell the compile of which kind of Scanner you are about to use, because the Scanner here is undefined by anyone.
After a import statement, you can use the class Scanner directly and the compiler will know about it.
Also, you can do this without using the import statement, although I don’t recommend:
java.util.Scanner scanner = new java.util.Scanner(System.in);
In this case, you just directly tell the compiler about which Scanner you mean to use.
answered May 11, 2011 at 4:08
lamwaiman1988lamwaiman1988
3,71715 gold badges55 silver badges87 bronze badges
0
You have to import java.util.Scanner at first line in the code
import java.util.Scanner;
answered May 11, 2011 at 4:00
Eng.FouadEng.Fouad
115k70 gold badges312 silver badges416 bronze badges
0
You need to include the line import java.util.Scanner;
in your source file somewhere, preferably at the top.
answered May 11, 2011 at 3:58
dlevdlev
48k5 gold badges125 silver badges132 bronze badges
You can resolve this error by importing the java.util.*
package — you can do this by adding following line of code to top of your program (with your other import
statements):
import java.util.*;
answered Jul 15, 2016 at 9:25
sometimes this can occur during when we try to print string from the user so before we print we have to use
eg:
Scanner scan=new Scanner (System.in);
scan.nextLine();
// if we have output before this string from the user like integer or other dat type in the buffer there is /n (new line) which skip our string so we use this line to print our string
String s=scan.nextLine();
System.out.println(s);
answered May 7, 2020 at 18:01
Please add the following line on top of your code
*import java.util.*;*
This should resolve the issue
Jim Simson
2,7643 gold badges22 silver badges30 bronze badges
answered Sep 5, 2020 at 12:58
Add import java.util.Scanner; at the very top of your code. Worked for me.
answered Dec 15, 2020 at 18:51
Во-первых, нажимай кнопку «Ответить» под комментарием, если хочешь, чтобы я получил уведомление.
Во-вторых, где ошибка? Ты видишь просто текстовое представление сканера, чего ты так перепугался-то? Ты до этого зачем-то выводил сканер в консоль, поэтому видел этот же текст в консоли. Тут никакой ошибки нет, тут написано, что это объект класса сканер, и какие у него характеристики. Или что ты там хотел увидеть, надпись «Здесь был Вася»?
Написал максимально легенькую программу, чтобы не захламлять вопрос:
package ****;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner (System.in);
System.out.println("Enter your name: ");
String name = sc.nextLine();
System.out.println("Enter your age: ");
int age = sc.nextInt();
}
}
Ошибка:
Task :run FAILED
Exception in thread «main» java.util.NoSuchElementException: No line found
at java.util.Scanner.nextLine(Scanner.java:1540)
at Expirement.Main.main(Main.java:7)
FAILURE: Build failed with an exception.
* What went wrong:
Execution failed for task ‘:run’.
> Process ‘command ‘C:Program FilesJavajdk1.8.0_271binjava.exe» finished with non-zero exit value 1
* Try:
Run with —stacktrace option to get the stack trace. Run with —info or —debug option to get more log output. Run with —scan to get full insights.
* Get more help at https://help.gradle.org
BUILD FAILED in 1s
2 actionable tasks: 2 executed
What is the Cannot Find Symbol Java Error?
Learn via video course
Java Course Online for Beginners
Tarun Luthra
Free
5
Enrolled: 50420
Start Learning
The Cannot Find Symbol in Java is a compilation error that occurs when we try to refer to something that is not present in the Java Symbol Table. A very common example of this error is using a variable that is not declared in the program.
Java compilers create and maintain Symbol tables. The symbol table stores the information of the identifiers i.e. classes, interfaces, variables, and methods in the given source of code. When we use these identifiers in our code, the compiler looks up to the symbol table for information. If the identifier is not declared or is not present in the given scope the compiler throws ‘Cannot Find Symbol’.
What Causes the Java Cannot Find Symbol Error?
There are multiple ways and reasons this error can occur, they all boil down to the fact that the online Java compiler couldn’t find the identifier in the Symbol table. Some of the very commonly made mistakes are listed below:
- Java is case sensitive, thus hello and Hello are two different variables.
- If identifiers are misspelled say the variable’s name is name and we are trying to access nmae can cause an error.
- Sometimes variables are declared in a particular iterating loop or method and if we try to call them from outside causes an error as it is out of the scope of the variable.
- Using variables without their declaration is also the common cause of the Java ‘Cannot find Symbol’ error.
- Sometimes we use methods from classes of a different package. To do so we are required to import the package of the class. Missing import statements can cause the error too.
- Use of underscore, dollar sign, numbers, letter even a character before or after different from the initial declaration can cause ‘Cannot Find Symbol’.
Examples of Java Cannot Find Symbol error
Let us now see these common errors in the form of code:
Example 1
Let us take a very simple example of a function that takes user input of a number and returns if the number is even or odd.
Output:
In the example above, the Scanner class is not imported and thus it is throwing a Cannot Find Symbol error. The compiler is not able to find Scanner in the program. Once we import the package, it is able to identify the Scanner class.
Example 2
Another example is wrong spelling and undeclared variables. Let us take a simple example where we multiply a variable a few times in a loop.
Output:
In the example above we can see 3 errors, all of them being ‘cannot find symbol’. The first one is because N is not declared in the program, and the two are because we declared finalResult but we are calling final_Result. Thus, a single underscore can also cause an error.
Example 3
Lastly, out-of-scope variables also cause a ‘Cannot find Symbol’ error, below example demonstrates it.
Output:
In the example above, the sum variable is declared inside the for loop and its scope is only inside for loop. As we are trying to access it outside the for loop, it is causing the error.
Structure of Java Cannot Find Symbol
Look at the output generated by the compiler for the 3rd example:
As we can see in the output, the compiler tells the line number of the error as well as points to the variable where the error occurred and the type of the error which is cannot find symbol in this case.
The compiler also produces the cannot find symbol including these two additional fields:
- The very first one is the symbol it is unable to find, i.e. the name and type of the referenced identifier.
- The other field is the location of the place where the symbol is used. The class and line number are returned where the identifier has been referenced.
Cannot Find Symbol vs Symbol Not Found vs Cannot Resolve Symbol
Different compilers may use slightly different terminologies. The ‘Cannot Find Symbol’, ‘Symbol Not Found’, and ‘Cannot Resolve Symbol’ are all same errors only. Besides the naming, these convey the same meaning to the user and the compiler.
Learn more about Java errors
Conclusion
- The ‘Cannot Find Symbol’ error in Java is a compilation error caused when the compiler cannot find a reference to an identifier.
- Silly mistakes like case sensitivity, use of underscore, use of undeclared variables, use of out-of-scope variables, etc. can cause this error.
- It is also identified by ‘Symbol Not Found’ and ‘Cannot Resolve Symbol’.
tick-icon22
30+ free video courses by top instructors at Scaler
tick-icon22
2000+ challenges to solve, curated by industry experts
tick-icon22
5+ masterclasses, monthly by experts on real time problem solving
tick-icon22
Fortnightly contests on DSA
tick-icon22
35+ topics library of different tech stacks.
2 Lakh + users already signed in to explore Scaler Topics!
There’s a common error that tends to stump new Java programmers. It happens when you group together a bunch of input prompts and one of the scanner.nextLine()
method calls gets skipped – without any signs of failure or error.
Take a look at the following code snippet, for example:
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("What's your name? ");
String name = scanner.nextLine();
System.out.printf("So %s. How old are you? ", name);
int age = scanner.nextInt();
System.out.printf("Cool! %d is a good age to start programming. nWhat language would you prefer? ", age);
String language = scanner.nextLine();
System.out.printf("Ah! %s is a solid programming language.", language);
scanner.close();
}
}
The first scanner.nextLine()
call prompts the user for their name. Then the scanner.nextInt()
call prompts the user for their age. The last scanner.nextLine()
call prompts the user for their preferred programming language. Finally, you close the scanner object and call it a day.
It’s very basic Java code involving a scanner
object to take input from the user, right? Let’s try to run the program and see what happens.
If you did run the program, you may have noticed that the program asks for the name, then the age, and then skips the last prompt for the preferred programming language and abruptly ends. That’s what we’re going to solve today.
Why Does the scanner.nextLine()
Call Get Skipped After the scanner.nextInt()
Call?
This behavior is not exclusive to just the scanner.nextInt()
method. If you call the scanner.nextLine()
method after any of the other scanner.nextWhatever()
methods, the program will skip that call.
Well, this has to do with how the two methods work. The first scanner.nextLine()
prompts the user for their name.
When the user inputs the name and presses enter, scanner.nextLine()
consumes the name and the enter or the newline character at the end.
Which means the input buffer is now empty. Then the scanner.nextInt()
prompts the user for their age. The user inputs the age and presses enter.
Unlike the scanner.nextLine()
method, the scanner.nextInt()
method only consumes the integer part and leaves the enter or newline character in the input buffer.
When the third scanner.nextLine()
is called, it finds the enter or newline character still existing in the input buffer, mistakes it as the input from the user, and returns immediately.
As you can see, like many real life problems, this is caused by misunderstanding between the user and the programmer.
There are two ways to solve this problem. You can either consume the newline character after the scanner.nextInt()
call takes place, or you can take all the inputs as strings and parse them to the correct data type later on.
How to Clear the Input Buffer After the scanner.nextInt()
Call Takes Place
It’s easier than you think. All you have to do is put an additional scanner.nextLine()
call after the scanner.nextInt()
call takes place.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("What's your name? ");
String name = scanner.nextLine();
System.out.printf("So %s. How old are you? ", name);
int age = scanner.nextInt();
// consumes the dangling newline character
scanner.nextLine();
System.out.printf("Cool! %d is a good age to start programming. nWhat language would you prefer? ", age);
String language = scanner.nextLine();
System.out.printf("Ah! %s is a solid programming language.", language);
scanner.close();
}
}
Although this solution works, you’ll have to add additional scanner.nextLine()
calls whenever you call any of the other methods. It’s fine for smaller programs but in larger ones, this can get very ugly very quick.
How to Parse Inputs Taken Using the scanner.nextLine()
Method
All the wrapper classes in Java contain methods for parsing string values. For example, the Integer.parseInt()
method can parse an integer value from a given string.
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
System.out.print("What's your name? ");
String name = scanner.nextLine();
System.out.printf("So %s. How old are you? ", name);
// parse the integer from the string
int age = Integer.parseInt(scanner.nextLine());
System.out.printf("Cool! %d is a good age to start programming. nWhat language would you prefer? ", age);
String language = scanner.nextLine();
System.out.printf("Ah! %s is a solid programming language.", language);
scanner.close();
}
}
This is a cleaner way of mixing multiple types of input prompts in Java. As long as you’re being careful about what the user is putting in, the parsing should be alright.
Conclusion
I’d like to thank you from the bottom of my heart for taking interest in my writing. I hope it has helped you in one way or another.
If it did, feel free to share with your connections. If you want to get in touch, I’m available on Twitter and LinkedIn.
Learn to code for free. freeCodeCamp’s open source curriculum has helped more than 40,000 people get jobs as developers. Get started