java - 依赖默认编码

Question

我正在使用 FindBugs 并且此错误不断生成:

Reliance on default encoding:

Found a call to a method which will perform a byte to String (or String to byte) conversion, and will assume that the default platform encoding is suitable. This will cause the application behavior to vary between platforms. Use an alternative API and specify a charset name or Charset object explicitly.

我认为这与扫描器有关，这是我的代码:

package mystack;

 import java.util.*;
  
  public class MyStack {
   
    private int maxSize;
    private int[] stackArray;
    private int top;
    
    public MyStack(int s) {
       maxSize = s;
       stackArray = new int[maxSize];
       top = -1;
    }
    public void push(int j) {
       stackArray[++top] = j;
    }
    public int pop() {
       return stackArray[top--];
    }
    public int peek() {
       return stackArray[top];
    }
    public int min() {
       return stackArray[0];
    }
    public boolean isEmpty() {
       return (top == -1);
    }
    public boolean isFull() {
       return (top == maxSize - 1);
    }
     
    
     public static void main(String[] args) {
         
       Scanner read = new Scanner(System.in);

         char end;
         
         System.out.println("Please enter the size of the Stack: ");
            int size=read.nextInt();
        
            MyStack stack = new MyStack(size);
         
         do{
         
            if(stack.isEmpty()){
              System.out.println("Please fill the Stack: (PUSH) \nBecause Stack is Empty.");
                 int add;
                 for(int i=0; i<size; i++)
                 {add=read.nextInt();
                   stack.push(add);}
                      }//End of if
        
          else if(stack.isFull()){
              System.out.println("Do you want to 1)POP 2)Know the Peek 3)Know the Min");
               int option=read.nextInt();
                if(option==1) 
                 stack.pop();
                
                else if (option==2)
                 System.out.println("The Peek= "+stack.peek());
                
                else if (option==3)
                 System.out.println("The Min= "+stack.min());
                
                else System.out.println("Error, Choose 1 or 2 or 3");
                      }//End of if
         else 
            { System.out.println("Do you want to 1)POP 2)Know the Peek 3)Know the Min 4)PUSH");
               int option=read.nextInt();
                 
                 if(option==1)
                     stack.pop();
                
                 else if (option==2)
                  System.out.println("The Peek= "+stack.peek());
                 
                 else if (option==3)
                  System.out.println("The Min= "+stack.min());
               
                 else if(option==4)
                   {int add=read.nextInt();
                    stack.push(add);}
                   }//end else
        
         System.out.print("Stack= ");
          for(int i=0; i<=stack.top; i++)
                 { System.out.print(stack.stackArray[i]+" ");}
         
         System.out.println();
         System.out.println();
         
         System.out.println("Repeat? (e=exit)");
          end=read.next().charAt(0);
          
         System.out.println();
         }while(end!='e');   
    System.out.println("End Of Program");
    }//end main

    }//end MyStack

它显然是一个堆栈，工作正常。

Answer 1

FindBugs 担心默认字符编码。如果你在 Windows 下，你的默认字符编码可能是“ISO-8859-1”。如果你在 Linux 下，它可能是“UTF-8”。如果您使用的是 MacOS，您可能正在使用“MacRoman”。您可能想阅读有关 charset encodings 的更多信息并在 available encodings in Java 上了解更多信息通过点击链接。

特别是，此行使用默认平台编码从控制台读取文本:

   Scanner read = new Scanner(System.in);

为确保代码在不同环境下工作相同，FindBugs 建议您将此更改为

   Scanner read = new Scanner(System.in, "UTF-8");

(或您最喜欢的编码)。这将保证，给定一个使用编码“UTF-8”的输入文件，无论您在哪台机器上执行程序，它都将以相同的方式被解析。

在您的情况下，您可以放心地忽略此警告，除非您有兴趣将文本文件输入到您的应用程序中。