Question

这段代码是否正确？我在某些地方以评论的形式提到了我的怀疑：

public class pract1
{
    public static void main (String[] args)
    {
        int i;
        String [] array = new String[20]; // Is this declaration correct?
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        System.out.println("Enter the Array: ");
        array = br.readLine(); // Is this the correct way to accept input from keyboard?
        i=0;
        while(array[i]!='\0') // Can I use the null pointer concept in Java?
        {
            System.out.println("The "+(i+1)+"character is:" +array[i]+"\n"); //Want to print each and every characters in string along with its position
            i++;
        }   
    }
}

Answer 1

Java字符串是一个对象，而不是数组。如果您对数组更熟悉，请使用split（）来获取String数组。您应该注意，派生数组中的每个元素都是String Object，只有一个字母，而不是原始Char。

String[] array = "abc".split("");
for(int i; i< array.length; i++)
{
     System.out.println("The "+(i+1)+"character is:" +array[i]+"\n"); 
}

您可能希望阅读本课题以了解Java Difference between "char" and "String" in Java

中的String Object

Answer 2

数组声明

String [] array = new String[20]; // Is this declaration correct?

此声明对于包含String类型的20个插槽的数组是正确的（所有这些插槽都初始化为null）。但你可能不会立即需要这个，或者不是这种形式。您不需要初始化变量，尤其是在您通过以下指令覆盖初始化时：array = ...。

获取输入

array = br.readLine(); // Is this the correct way to accept input from keyboard?

是的，readLine()是要走的路，但是as the doc states，正如编译器告诉你的那样，readLine();不会返回String的数组但是一个String。您应该使用String变量：

// initialize a String variable 'line' containing the whole line without the '\n' at the end
String line = br.readLine();

更新：您也可以使用the Scanner class，而这通常就是我们所做的。然后，使用sc.nextLine()获得与br.readLine()，as stated there类似的结果。

打印字符

System.out.println("The "+(i+1)+"character is:" +array[i]+"\n");
//Want to print each and every characters in string along with its position

您可以通过方法charAt(int)访问String中给定位置的字符。此外，您不必处理复杂的C内容，例如通过'\0'查找字符串的结尾。您应该使用正确的for循环，如下所示：

String line = br.readLine();
for (int i = 0; i < line.length(); i++) {
    System.out.println("The "+(i+1)+" character is: " + line.charAt(i) +"\n");
}

另一种解决方案是使用the toCharArray() method：

String line = br.readLine();
char[] chars = line.toCharArray();
for (int i = 0; i < chars.length; i++) {
    System.out.println("The "+(i+1)+" character is: " + chars[i] +"\n");
}

Answer 3

你也可以这样做：

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class Pract1
{
    public static void main (String[] args)
    {
        String array = ""; //initialize an empty string that will hold the value from the keyboard
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); // initialize the BufferedReader

        // while "stop" is not typed on the console, keep running
        while(!array.equals("stop")) {
            System.out.println("Enter the Array: ");
            try {
                array = br.readLine(); // read from the console
                System.out.println("Array:" + array);
                for(int i=0; i<array.length(); i++) { //loop through the input and show the chars
                    System.out.println("Character "+(i+1)+" is: " +array.charAt(i));
                }

            } catch (Exception e) { // catch any exception and show only the message, not the entire stackTrace
                System.out.println("Error: " +  e.getMessage());
            }
            System.out.println("");
        }

    }
}

Java中的字符串操作

3 个答案:

数组声明

获取输入

打印字符