显式转换为相同类型

Question

显式转换为相同类型是否会影响性能，还是它们会被编译器过滤而永远不会到达字节码？

例：

int x = 3;
int y = (int) x;

Answer 1

在此类上运行javap -c：

public class SameTypeCastsDemo {

    public static void withoutCasts() {
        int x = 2;
        int y = x;
        System.out.println(y);
    }

    public static void withCast() {
        int x = 2;
        int y = (int) x;
        System.out.println(y);
    }

}

显示字节码看起来相同：

public static void withoutCasts();
  Code:
   0:   iconst_2
   1:   istore_0
   2:   iload_0
   3:   istore_1
   4:   getstatic   #2; //Field java/lang/System.out:Ljava/io/PrintStream;
   7:   iload_1
   8:   invokevirtual   #3; //Method java/io/PrintStream.println:(I)V
   11:  return

public static void withCast();
  Code:
   0:   iconst_2
   1:   istore_0
   2:   iload_0
   3:   istore_1
   4:   getstatic   #2; //Field java/lang/System.out:Ljava/io/PrintStream;
   7:   iload_1
   8:   invokevirtual   #3; //Method java/io/PrintStream.println:(I)V
   11:  return

更新：使用非原始对象引用：

public class SameTypeCastsDemo {
    Integer x;
    Integer y;

    public SameTypeCastsDemo(Integer x, Integer y) {
        this.x = x;
        this.y = y;
    }

    public void print() {
        System.out.println(y);
    }

    public static void withoutCasts() {
        SameTypeCastsDemo x = new SameTypeCastsDemo(2, 3);
        SameTypeCastsDemo y = x;
        y.print();
    }

    public static void withCast() {
        SameTypeCastsDemo x = new SameTypeCastsDemo(2, 3);
        SameTypeCastsDemo y = (SameTypeCastsDemo) x;
        y.print();
    }

}

javap -c SameTypeCastsDemo：

public static void withoutCasts();
  Code:
   0:   new #6; //class SameTypeCastsDemo
   3:   dup
   4:   iconst_2
   5:   invokestatic    #7; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
   8:   iconst_3
   9:   invokestatic    #7; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
   12:  invokespecial   #8; //Method "<init>":(Ljava/lang/Integer;Ljava/lang/Integer;)V
   15:  astore_0
   16:  aload_0
   17:  astore_1
   18:  aload_1
   19:  invokevirtual   #9; //Method print:()V
   22:  return

public static void withCast();
  Code:
   0:   new #6; //class SameTypeCastsDemo
   3:   dup
   4:   iconst_2
   5:   invokestatic    #7; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
   8:   iconst_3
   9:   invokestatic    #7; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
   12:  invokespecial   #8; //Method "<init>":(Ljava/lang/Integer;Ljava/lang/Integer;)V
   15:  astore_0
   16:  aload_0
   17:  astore_1
   18:  aload_1
   19:  invokevirtual   #9; //Method print:()V
   22:  return

Answer 2

Sun将其称为身份转换。

-引用链接-

任何类型都允许从类型到相同类型的转换。

这看似微不足道，但有两个实际后果。 首先，始终允许表达式以所需的类型开头，从而允许简单陈述的规则，即如果只是琐碎的标识转换，则每个表达式都必须进行转换。 其次，它意味着为清楚起见，允许程序包含冗余的强制转换运算符。