如何判断整数只有2位数字？

Question

I need to write a Java program that prompts the user to enter an integer consisting of exactly 2 digits; then displays on the screen the sum of its individual digits.

I am stuck here. What am I doing wrong?

import java.util.Scanner ;

public class ss {
    public static void main(String[] args)
    {

       Scanner input  = new Scanner(System.in);

       int x;

       System.out.println("Please Enter a number consists of 2 digits only : ");

       x = input.nextInt();
       x.length() == 2;
   }
}

and the last line contains an error!

Answer 1

Assuming that x is positive, a simple way to check if it has exactly two digits would be:

if (x >= 10 && x <= 99) {
    // x contains exactly two digits
}

Answer 2

The variable x is of type int , so you can't call a method on it. You need to either read the input as a String or convert the int to a String then call length() , or just test that the int is between 10 and 99 , inclusive.

Answer 3

In a programming langauge, there are things called L-values and R-values. In an assignment operation, a L-value can accept a R-value as input. This comes from the typical layout which has L-values on the left of the assignment operator and R-values on the right side of the assignment operator.

x = 5;

x is the L-value and 5 is the R-value. It is possible to assign five to x.

However, a function returns a R-value. Therefore, it is possible to do this

x = a.length();

but is is not possible to do

a.length() = x;

because you can not assign a value to the return of a function.

Fundamentally, L-values are names which represent a value , but R-values are values or items which when analyzed result in the return of values .

Now, if you used the equals comparison operator, both values must be R-values, because no assignment is being performed

a.length == x

is just fine, because it is not the assignment operator = but rather one of the comparison operators == .

Answer 4

Your error comes because x is a primitive, not an object. Only objects have methods like length() . A quick an easy way to determine the length of an integer is by using Math.log() .

public int length(int n){
    if (n == 0) return 1; // because Math.log(0) is undefined
    if (n < 0) n = -n; // because Math.log() doesn't work for negative numbers
    return (int)(Math.log10(n)) + 1; //+1 because Math.log10 returns one less
                                     //than wanted. Math.log10(10) == 1.
}

This method uses the fact that the base b logarithm of an integer a is related to the length of the integer a.

Or, if you don't know how to use methods, you could do this (assuming n is the integer to check):

int length = (n == 0)? 1: ((n > 0)? (int) (Math.log(n)) + 1: (int) (Math.log(-n)) + 1);

Or, if you don't use the ternary operator, you could expand it:

int length = -1; //placeholder; might not need it.
if (n == 0) length = 1;
else if (n > 0) length = (int) (Math.log(n)) + 1;
else length = (int) (Math.log(-n)) + 1;

Answer 5

You can't find the length of an int by calling a method on it, but you can find the length of a String .

Try converting the int to a String and finding the length of that:

boolean isTwoDigits = x.toString().length() == 2;

Answer 6

You cannot call length on integer just write

if(x>=10 && x<=99)
{
//write your code here
}