自定義網絡庫和序列化VS默認序列化和RMI

Question

我有這個類，我通過RMI發送這個類，並通過kryonet和kryo，我得到的rmi兩個10個對象的數組作爲遠程方法的返回打電話，如果kryonet從服務器到客戶端使用kryonet和kryo回顯，我總共有一個對象55 * 10其中555但RMI爲1196bytes，

這些結果是否合理？

*爲什麼這些結果是這樣的？

爲什麼會有這麼多的差異。

這過頂球或其他因素參與幕後，

這些都使得在RMI那麼多總和太多的區別，並指出了我。

而且單個對象的總共55個字節是可以嗎？*。

我只是需要一些確認和專家的眼睛，因爲我有呈現結果，

我會真的很感謝。

這是類我米既使用：

public class TBall { 

private float x, y; // Ball's center (x, y) 
private float speedX, speedY; // Ball's speed per step in x and y 
private float radius; // Ball's radius 
private Color color; // Ball's color 

public boolean collisionDetected = false; 
public static boolean run = false; 

private String name; 

private float nextX, nextY; 
private float nextSpeedX, nextSpeedY; 

public TBall() { 
    super(); 
} 

public TBall(String name1, float x, float y, float radius, float speed, 
     float angleInDegree, Color color) { 
    this.x = x; 
    this.y = y; 
    // Convert velocity from polar to rectangular x and y. 
    this.speedX = speed * (float) Math.cos(Math.toRadians(angleInDegree)); 
    this.speedY = speed * (float) Math.sin(Math.toRadians(angleInDegree)); 
    this.radius = radius; 
    this.color = color; 
    this.name = name1; 
} 

public String getName() { 
    return this.name; 
} 

public float getSpeed() { 
    return (float) Math.sqrt(speedX * speedX + speedY * speedY); 
} 

public float getMoveAngle() { 
    return (float) Math.toDegrees(Math.atan2(speedY, speedX)); 
} 

public float getRadius() { 
    return radius; 
} 

public Color getColor() { 
    return this.color; 
} 

public void setColor(Color col) { 
    this.color = col; 
} 

public float getX() { 
    return x; 
} 

public float getY() { 
    return y; 
} 

public void setX(float f) { 
    x = (int) f; 
} 

public void setY(float f) { 
    y = (int) f; 
} 

public void move() { 
    if (collisionDetected) { 
     // Collision detected, use the values computed. 
     x = nextX; 
     y = nextY; 
     speedX = nextSpeedX; 
     speedY = nextSpeedY; 
    } else { 
     // No collision, move one step and no change in speed. 
     x += speedX; 
     y += speedY; 
    } 
    collisionDetected = false; // Clear the flag for the next step 

    System.out.println("In serializedBall in move."); 
} 

public void collideWith() { 

    float minX = 0 + radius; 
    float minY = 0 + radius; 
    float maxX = 0 + 640 - 1 - radius; 
    float maxY = 0 + 480 - 1 - radius; 

    double gravAmount = 0.9811111f; 
    double gravDir = (90/57.2960285258); 

    // Try moving one full step 
    nextX = x + speedX; 
    nextY = y + speedY; 
    System.out.println("In serializedBall in collision."); 

    // If collision detected. Reflect on the x or/and y axis 
    // and place the ball at the point of impact. 
    if (speedX != 0) { 
     if (nextX > maxX) { // Check maximum-X bound 
      collisionDetected = true; 
      nextSpeedX = -speedX; // Reflect 
      nextSpeedY = speedY; // Same 
      nextX = maxX; 
      nextY = (maxX - x) * speedY/speedX + y; // speedX non-zero 
     } else if (nextX < minX) { // Check minimum-X bound 
      collisionDetected = true; 
      nextSpeedX = -speedX; // Reflect 
      nextSpeedY = speedY; // Same 
      nextX = minX; 
      nextY = (minX - x) * speedY/speedX + y; // speedX non-zero 
     } 
    } 
    // In case the ball runs over both the borders. 
    if (speedY != 0) { 
     if (nextY > maxY) { // Check maximum-Y bound 
      collisionDetected = true; 
      nextSpeedX = speedX; // Same 
      nextSpeedY = -speedY; // Reflect 
      nextY = maxY; 
      nextX = (maxY - y) * speedX/speedY + x; // speedY non-zero 
     } else if (nextY < minY) { // Check minimum-Y bound 
      collisionDetected = true; 
      nextSpeedX = speedX; // Same 
      nextSpeedY = -speedY; // Reflect 
      nextY = minY; 
      nextX = (minY - y) * speedX/speedY + x; // speedY non-zero 
     } 
    } 

    System.out.println("In serializedBall collision."); 
    // speedX += Math.cos(gravDir) * gravAmount; 
    // speedY += Math.sin(gravDir) * gravAmount; 

    System.out.println("In serializedBall in collision."); 
} 

}

感謝。

Answer 1

你從哪裏得到'55'？您有：

9浮筒，= 9X4個字節，總共36個字節 1布爾值，序列化爲一個字節，總共1個字節 1字符串，可以是任何長度 1顏色，又包含： 1 INT，序列化爲4個字節 1個浮子，序列化爲4個字節 2浮[]長度3各自，序列化爲24個字節 1彩色空間，而這又包含： 2整數，序列化爲8個字節

總的這至少需要77個字節加上傳輸字符串所需的任何東西。

序列化還會在每個項目前發送類信息，版本控制信息和標籤; RMI也發送方法信息。所有這些都可以輕鬆解釋這種差異。我不知道那些其他軟件包是做什麼的。