我怎么能阻止这个整数溢出?

时间:2012-08-09 01:23:02

标签: java int long-integer

好吧,所以我遇到了这个问题,你可能会因为实际支付而得到更多的故障。这很难解释,当你增加金额时,首先这是我的代码:

    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);
    }

总价格是溢出的,但我不确定如何让它停止。有人建议使用这样的东西:

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

但我不知道如何在此代码中使用它。有人可以帮帮我吗?

3 个答案:

答案 0 :(得分:2)

答案 1 :(得分:0)

long totalPrice = price * amount;

priceamount投射到long,以便进行长时间的计算:

long totalPrice = (long)price * amount;

答案 2 :(得分:0)

将您的总数初始化为零...类似程序。转到http://codejava.co.uk/download_code.html