Java 我怎么知道一个整数只有 2 位数长？
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How can I tell that an integer is only 2 digits long?
提问by whoseee
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.
我需要编写一个 Java 程序，提示用户输入一个由 2 位数字组成的整数；然后在屏幕上显示其各个数字的总和。
I am stuck here. What am I doing wrong?
我被困在这里。我究竟做错了什么？
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!
最后一行包含错误！
回答by óscar López
Assuming that xis positive, a simple way to check if it has exactly two digits would be:
假设这x是肯定的，检查它是否正好有两位数的简单方法是：
if (x >= 10 && x <= 99) {
// x contains exactly two digits
}
回答by rgettman
The variable xis of type int, so you can't call a method on it. You need to either read the input as a Stringor convert the intto a Stringthen call length(), or just test that the intis between 10and 99, inclusive.
该变量x的类型为int，因此您不能对其调用方法。您需要将输入读取为 aString或将 the 转换int为Stringthen call length()，或者只是测试int介于10和之间99，包括。
回答by Edwin Buck
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.
在编程语言中，有称为 L 值和 R 值的东西。在赋值操作中，L 值可以接受 R 值作为输入。这来自典型的布局，其中 L 值位于赋值运算符的左侧，而 R 值位于赋值运算符的右侧。
x = 5;
x is the L-value and 5 is the R-value. It is possible to assign five to x.
x 是 L 值，5 是 R 值。可以将 5 分配给 x。
However, a function returns a R-value. Therefore, it is possible to do this
但是，函数返回 R 值。因此，可以做到这一点
x = a.length();
but is is not possible to do
但这是不可能的
a.length() = x;
because you can not assign a value to the return of a function.
因为你不能为函数的返回值赋值。
Fundamentally, L-values are nameswhich represent a value, but R-values are valuesor items which when analyzed result in the return of values.
从根本上说，L值是名称，其代表一个值，但R值是值，其在返回解析结果时或项目值。
Now, if you used the equals comparison operator, both values must be R-values, because no assignment is being performed
现在，如果您使用等于比较运算符，则两个值都必须是 R 值，因为没有执行赋值
a.length == x
is just fine, because it is not the assignment operator =but rather one of the comparison operators ==.
很好，因为它不是赋值运算符=，而是比较运算符之一==。
回答by Justin
Your error comes because xis 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().
你的错误是因为它x是一个原始的，而不是一个对象。只有对象具有像length(). 确定整数长度的一种快速简便的方法是使用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.
该方法利用了一个事实，即整数 a 的底 b 对数与整数 a 的长度有关。
Or, if you don't know how to use methods, you could do this (assuming n is the integer to check):
或者，如果您不知道如何使用方法，您可以这样做（假设 n 是要检查的整数）：
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:
或者，如果您不使用三元运算符，则可以扩展它：
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;
回答by earora4498
You can't find the length of an intby calling a method on it, but you can find the length of a String.
您无法int通过对其调用方法来找到an 的长度，但是您可以找到 a 的长度String。
Try converting the intto a Stringand finding the length of that:
尝试将inta转换为 aString并找到它的长度：
boolean isTwoDigits = x.toString().length() == 2;
回答by Gourav
You cannot call length on integer just write
你不能在整数上调用 length 只写
if(x>=10 && x<=99)
{
//write your code here
}
