java - 如何捕获 NumberFormatException？

Question

我正在尝试找出如何在代码中捕获数字格式异常错误，这样如果用户在字符串中输入字母并且我的程序尝试将其解析为 int 我的程序将不会抛出错误而是停止并返回一个 boolean 值。我还试图了解，如果 try 语句有效，我希望它继续执行以下代码。

    if (counter == 3) {
        int compare;
        boolean check = true;
        String[] newip = IpAddress.split("\\.");
        if (newip.length == 4) {
            for (int index = 0; index < newip.length; index++) {
                //There should be a try statement here.
                // if the try statement fails then I'd like for it to catch
                // the numberformatexception and evaluate my boolean to 
                //false;
                //but if it passes I'd like for it to continue to execute 
                //the following code.
                    compare = Integer.parseInt(newip[index]);
                if (compare >= 0 & (compare <= 255)) {
                    check = true;
                }
                else{
                    check = false;
                }
            }
            if (check)
                return true;
            else
                return false;
        }
        else {
            check = false;
            return check;
        }
    }
    else{
        return false;
    }
}

Answer 1

用 try/catch 将该行括起来:

try {
    compare = Integer.parseInt(newip[index]);
} catch (NumberFormatException e) {
    check = false;
}

然后:

if (check) {
    if (compare >= 0 & (compare <= 255)) {
        check = true;
    } else {
        check = false;
    }
} else {
    return false;
}