Actionscript 3.0 Splice/RemoveChild問題

我有岩石和敵人忍者離開屏幕。忍者一旦遭到手裏劍的擊中就應該消失，然而他們正在接受兩次命中消失，儘管手中的每一次都會消失。代碼對於shurikens和敵人忍者幾乎是相同的，但是忍者似乎不能正常工作。我也偶爾忍受忍者卡在屏幕上的任何地方，手中穿過它。Actionscript 3.0 Splice/RemoveChild問題

//If a rock moves off the screen, it is spliced. 
    if (rock[j].x <= -301){ 
     removeChild(rock[j]); 
     rock.splice(j, 1); 
     rockNum--; 
    } 
} 

for (var q=shurikens.length - 1; q >= 0; q--){ 
    for (var w=enemy.length - 1; w >= 0; w--){ 
     //If the shurikens hit the ninjas. 
     if ((shurikens[q]).hitTestObject(enemy[w])){ 
      removeChild(enemy[w]); 
      enemy.splice(w, 1); 
      ninjaNum--; 
      removeChild(shurikens[q]); 
      shurikens.splice(q, 1); 
      break; 
     } 
    } 
}

}

來源

2017-08-29 Marcoo Poloo

你可以用兩個忍者手裏劍 - 檢查如果一個帶有索引「q」的手裏劍同時擊中兩個忍者會發生什麼。你會得到索引'q + 1'拼接出來的手裏劍，而不是從屏幕上移除。爲了解決這個問題，當你發現一個給定的手裏劍時，你需要立即停止外圈。 – Vesper

如何解決你的代碼，並給你一些性能提示：

這段代碼是非常「蟲友」：要修改陣列的「長度」屬性在依賴於相同屬性的for/in循環中，這實際上不是明智的做法。的方式我會做：

// Create an array where to store colliding ninjas and shurikens 
    var depleted:Array = []; 
    // Create local references to make code more readable and fast (querying arrays all the time is slow). 
    var ninja:DisplayObject; 
    var shuriken:DisplayObject; 
    // Loop in shurikens (no need to check if shurikens have been depleted, since you query only once each of them 
    for (var q:int=shurikens.length - 1; q >= 0; q--){ 
     shuriken = shurikens[q]; // Assign a shuriken to our variable so you avoid to call further shurikens[q]; 
     for (var w:int=enemy.length - 1; w >= 0; w--){ // Loop in ninjas 
      ninja = enemy[w]; // Assign ninja 
      //If the shurikens hit the ninjas. (only if ninjas have not yet been removed) 
      // This is the core of our improvement, before calling hitTest that is slow, you first check that ninjas have not already been killed 
      if (depleted.indexOf(ninja) == -1 && 
shuriken.hitTestObject(ninja)){ 
       // It's a hit with a live ninja. I just add both objects to depleted list. 
       depleted.push(ninja); 
       depleted.push(shuriken); 
       break; // Breaking the loop, makes sure shuriken cannot hit 2 ninjas 
      } 
     } 
    } 
// Then, you loop in the list of killed ninjas and depleted shurikens, and remove them from arrays and display list 
for each (var depletedObj:DisplayObject in depleted) { 
    // First remove object from the relevant array 
    if (shurikens.indexOf(depletedObj) != -1) shurikens.splice(shurikens.indexOf(depletedObj), 1); // If it was in the shurikens array remove from there 
    else if (enemy.indexOf(depletedObj) != -1) enemy.splice(enemy.indexOf(depletedObj), 1); // If it was in the ninjas array remove from there 
    // The do all necessary stuff to remove object from DisplayList (end eventually add it to an object pooling list) 
    removeChild(depletedObj); 
} 
ninjaNum = enemy.length; // Update number of ninjas

另一個提示，如果你對每一幀的循環運行它，如果你把忍者和shurikens在2個不同的DisplayObjectContainer，可以先則hitTest的2個大容器，和一旦碰撞，你可以運行循環來檢查碰撞。另外，在/ in循環的數字中，聲明變量始終爲：int。鍵入變量使訪問速度快於無類型變量。你當然可以改進這個代碼，使其更快，即：爲shurikens和ninjas添加一個「alive = true」屬性，所以你不需要查詢第三個數組等。

來源

2017-08-31 20:16:35

Actionscript 3.0 Splice/RemoveChild問題

回答

相關問題