我該如何阻止這個整數溢出？

Question

好的，所以我有這個問題，你可以在那裏得到超過你實際支付的小故障。這是一種很難解釋，當你加量，那麼首先這裏是我的代碼：

public void setAmount(Player player, int button) { 
     int amount = (Integer) player.getTemporaryAttribute("geAmount"); 
     int id = (Integer) player.getTemporaryAttribute("geItem"); 
     int price = (Integer) player.getTemporaryAttribute("price"); 
     long totalPrice = price * amount; 
     switch(button) { 
     case 157: 
      if(player.getRights() > 1){ 
       player.sm(""+totalPrice+""); 
      } 
      if(amount > 1) { 
       amount--; 
      } else { 
       amount = 1; 
      } 
      break; 
     case 159: 
      if(player.getRights() > 1){ 
       player.sm(""+totalPrice+""); 
      } 
      if(amount < Integer.MAX_VALUE && totalPrice <= Integer.MAX_VALUE && totalPrice >= 0) { 
       amount++; 
      } 
      break; 
     case 162: 
      if(player.getRights() > 1){ 
       player.sm(""+totalPrice+""); 
      } 
      if(player.getTemporaryAttribute("buying") == Boolean.TRUE && totalPrice <= Integer.MAX_VALUE && totalPrice >= 0) { 
       amount += 1; 
      } else { 
       amount = 1; 
      } 
      break; 
     case 164: 
      if(player.getRights() > 1){ 
       player.sm(""+totalPrice+""); 
      } 
      if(player.getTemporaryAttribute("buying") == Boolean.TRUE && totalPrice <= Integer.MAX_VALUE && totalPrice >= 0) { 
       amount += 10; 
      } else { 
       amount = 10; 
      } 
      break; 
     case 166: 
      if(player.getRights() > 1){ 
       player.sm(""+totalPrice+""); 
      } 
      if(player.getTemporaryAttribute("buying") == Boolean.TRUE && totalPrice <= Integer.MAX_VALUE && totalPrice >= 0) { 
       amount += 100; 
      } else { 
       amount = 100; 
      } 
      break; 
     case 168: 
      if(player.getRights() > 1){ 
       player.sm(""+totalPrice+""); 
      } 
      if(player.getTemporaryAttribute("buying") == Boolean.TRUE && totalPrice <= Integer.MAX_VALUE && totalPrice >= 0) { 
       amount += 1000; 
      } else { 
       amount = player.getInventory().getContainer().getNumberOff(id); 
      } 
      break; 
     case 171: 
     case 173: 
      if(player.getRights() > 1){ 
       player.sm(""+totalPrice+""); 
      } 
      player.getActionSender().sendConfig(1111, (Integer) player.getTemporaryAttribute("price")); 
      break; 
     } 
     if(amount == 0){ 
      amount = 1; 
     } 
     if(amount >= Integer.MAX_VALUE || amount < 0) { 
      amount = Integer.MAX_VALUE; 
     } 
     player.setTemporaryAttribute("geAmount", amount); 
     player.getActionSender().sendConfig(1109, id); 
     player.getActionSender().sendConfig(1110, amount); 
    } 

總價是什麼讓四溢，但我不確定如何讓它停下來。有人曾用這樣的建議：

int last = 0; 
int current = 0; 
while(last < current) //loop terminates on overflow 
    last = current++; 

但我不知道我會怎麼用，在該代碼。任何人都可以幫助我嗎？

Answer 1

long totalPrice = price * amount; 

演員price或amount到long所以計算作爲一項長期進行：

long totalPrice = (long)price * amount; 

Answer 2

使用BigInteger類

http://docs.oracle.com/javase/1.5.0/docs/api/java/math/BigInteger.html

Answer 3

你的總初始化爲零.. 。類似的程序。轉到http://codejava.co.uk/download_code.html