java - 即使正确识别了每个字符，使用 junit 比较字符串也不会断言为 true

Question

这个问题已经有答案了:

How do I compare strings in Java? (23 个回答)

已关闭 9 年前。

我不明白为什么我的单元测试失败了，

assertTrue( (String) temp=="c+(d/f)-e");

即使我连续打印 temp 的内容，它也是一样的。

这是代码

单元测试

import static org.junit.Assert.*;

import org.junit.Test;


public class LexerTest {


    @Test
    public void testGetInExpression(){
        //------
        Lexer lex=new Lexer("a+b*(c+(d/f)-e)-g");
        //--This test passes
        assertTrue((String)lex.getLexStr()=="a+b*(c+(d/f)-e)-g");
        //------

        for(int i=0;i<5;i++){
            lex.getNext();
        }

        //----------------
        //Method returns a in-expression (or a substring within first parantheses)
        Lexer sub=lex.getInExpression(); 
        String temp=sub.getLexStr(); //returns a type String
        System.out.println(temp);
        System.out.println(temp.length());

        //printing out each character of 'temp'
        for(int i=0;i<temp.length();i++){
            System.out.print(i);
            System.out.print(" = ");
            System.out.println(temp.charAt(i));
            //prints: the string 
            // 0 = c
            // 1 = +
            // ...
            // ..
            // 8 = e
        }

        //The following fails (mysteriously!!)
        assertTrue( (String) temp=="c+(d/f)-e");
        //assertTrue( (Character)lex.getNext()=='-');
    }
}

这是 Lexer 类:

public class Lexer {
    private int size;
    private int posn;
    private String lexStr;

    public Lexer(String val){
        lexStr=val;
        size=lexStr.length();
        posn=0;     
    }

    public String getLexStr(){
        return lexStr;
    }
    public Object getNext(){
        posn=posn+1;
        if(posn<size){return current();} //return character at next site
        else {return null;}              //return a null value to indicate end
    };

    public Object current(){
        if(posn<size){
            char c=lexStr.charAt(posn);
            if(Character.isDigit(c)){
                return Character.getNumericValue(c); //return an Integer object
            }else
                return c; //Return a plain character
        }
        else{
            return null;
        }
    };

    public Lexer getInExpression(){
        //This must return a subExpression in the original Expression enclosed by braces
            int open=1;
            int closed=0;
            String returnStr=null;
            String temp=lexStr.substring(posn);
            for(int a=0; a <temp.length(); a++){
                getNext();
                if(temp.charAt(a)=='('){
                    open++;
                }else if(temp.charAt(a)==')'){
                    closed++;
                    if(open==closed){
                        returnStr=temp.substring(0,a);
                        break;
                    }
                }
            }
            //---check for validity of returnStr
            if (returnStr==null){
                System.out.println("I cannot be null! Please check here");
                return null;
            }else{
                Lexer inExpr=new Lexer(returnStr);      
                return inExpr;
            }


    }
}

Answer 1

您应该使用assertEquals，而不是assertTrue。对于不同的 String 对象，使用 == 进行比较并不好。