为什么字符串比较不同？

第一个方法返回false，第二个返回true。我的想法：在 test1() 中，当创建 d 时 - 字符串池中已经存在 c。但事实证明 d 是重复的。test2() 的行为对我来说很清楚。你能解释一下在 test1() 中创建 c 和 d 背后的逻辑吗？

public class Main
{
    public static void main(String[] args) {
        System.out.println("Hello World");
        test1();
        test2();
    }
    
    public static void test1() {
        String a = "a";
        String b = "b";
        // String c = a + b;
        String c = a.concat(b);
        String d = "ab";
        System.out.println(c == d);
    }
    public static void test2() {
        String a = "ab";
        String b = "a" + "b";
        System.out.println(a == b);   
    }
}

字节码：

public static void test1();
    Code:
       0: ldc           #7                  // String a
       2: astore_0
       3: ldc           #8                  // String b
       5: astore_1
       6: aload_0
       7: aload_1
       8: invokevirtual #9                  // Method java/lang/String.concat:(Ljava/lang/String;)Ljava/lang/String;
      11: astore_2
      12: ldc           #10                 // String ab
      14: astore_3
      15: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
      18: aload_2
      19: aload_3
      20: if_acmpne     27
      23: iconst_1
      24: goto          28
      27: iconst_0
      28: invokevirtual #11                 // Method java/io/PrintStream.println:(Z)V
      31: return

  public static void test2();
    Code:
       0: ldc           #10                 // String ab
       2: astore_0
       3: ldc           #10                 // String ab
       5: astore_1
       6: getstatic     #2                  // Field java/lang/System.out:Ljava/io/PrintStream;
       9: aload_0
      10: aload_1
      11: if_acmpne     18
      14: iconst_1
      15: goto          19
      18: iconst_0
      19: invokevirtual #11                 // Method java/io/PrintStream.println:(Z)V
      22: return