如何在不使用 Integer 类的情况下在 Java 中获取 Integer.MAX_VALUE 的值
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How to get the value of Integer.MAX_VALUE in Java without using the Integer class
提问by madsword19
I have this question that has completely stumped me. I have to create a variable that equals Integer.MAX_VALUE... (in Java)
我有这个问题，完全难倒了我。我必须创建一个等于 Integer.MAX_VALUE... 的变量（在 Java 中）
// The answer must contain balanced parentesis
public class Exercise{
public static void main(String [] arg){
[???]
assert (Integer.MAX_VALUE==i);
}
}
The challenge is that the source code cannot contain the words "Integer", "Float", "Double" or any digits (0 - 9).
挑战在于源代码不能包含单词“Integer”、“Float”、“Double”或任何数字（0 - 9）。
回答by Hypothetical inthe Clavicle
As others have said.
正如其他人所说。
int i = Integer.MAX_VALUE;
is what you want.
是你想要的。
Integer.MAX_VALUE, is a "static constant" inside of the "wrapper class" Integer that is simply the max value. Many classes have static constants in them that are helpful.
Integer.MAX_VALUE，是“包装类”Integer 中的“静态常量”，它只是最大值。许多类中都有有用的静态常量。
回答by MrLore
The issue is that the answer cannot contain: "Integer", "Float", "Double", and digits (0 - 9)
问题是答案不能包含：“Integer”、“Float”、“Double”和数字 (0 - 9)
There are other things in Java which can be represented as an Integer, for example a char:
Java 中还有其他东西可以表示为 a Integer，例如 a char：
char aCharacter = 'a';
int asInt = (int) aCharacter;
System.out.println(asInt); //Output: 97
You can also add chars together in this manner:
您还可以通过这种方式将字符添加在一起：
char aCharacter = 'a';
char anotherCharacter = 'b';
int sumOfCharacters = aCharacter + anotherCharacter;
System.out.println(sumOfCharacters); //Output: 195
With this information, you should be able to work out how to get to 2147483647on your own.
有了这些信息，您应该能够弄清楚如何2147483647自己去。
回答by Críostóir ó Catháin
OK, so an Integer can only take certain values. This is from MIN_VALUE to MAX_VALUE where the minimum value is negative.
好的，所以 Integer 只能取某些值。这是从 MIN_VALUE 到 MAX_VALUE，其中最小值为负。
If you increase an integer past this upper bound the value will wrap around and become the lowest value possible. e.g. MAX_VALUE+1 = MIN_VALUE.
如果您增加一个整数超过此上限，该值将环绕并成为可能的最低值。例如 MAX_VALUE+1 = MIN_VALUE。
Equally, if you decrease an integer past the lower bound it will wrap around and become the largest possible value. e.g. MIN_VALUE-1 = MAX_VALUE.
同样，如果您将一个整数减小到超过下限，它将环绕并成为可能的最大值。例如 MIN_VALUE-1 = MAX_VALUE。
Therefore a simple program that instantiates an int, decrements it until it wraps around and returns that value should give you the same value as Integer.MAX_VALUE
因此，一个简单的程序实例化一个 int，递减它直到它环绕并返回该值应该给你相同的值 Integer.MAX_VALUE
public static void main(String [] arg) {
int i = -1
while (i<0) {
i--;
}
System.out.println(i);
}
回答by Stephen C
Here's a solution:
这是一个解决方案：
int ONE = "X".length();
int max = ONE;
while (max < max + ONE) {
max = max + ONE;
}
or lots of variants.
或很多变种。
(The trick you were missing is how to "create" an integer value without using a numeric literal or a number wrapper class. Once you have created ONE, the rest is simple ...)
（您缺少的技巧是如何在不使用数字文字或数字包装类的情况下“创建”一个整数值。一旦您创建了ONE，剩下的就很简单了……）
回答by Bohemian
Here's a succinct method:
这是一个简洁的方法：
int ONE = "x".length();
int i = -ONE >>> ONE; //unsigned shift
This works because the max integer value in binary is all ones, except the top (sign) bit, which is zero. But -1 in twos compliment binary is all ones, so by bit shifting -1 one bit to the right, you get the max value.
这是有效的，因为二进制中的最大整数值都是 1，除了最高（符号）位为零。但是 -1 在二进制补码二进制中都是 1，因此通过将 -1 向右移动一位，您可以获得最大值。
11111111111111111111111111111111 // -1 in twos compliment
01111111111111111111111111111111 // max int (2147483647)
回答by slyaer
A bit late, but here goes:
有点晚了，但这里是：
int two = "xx".length();
int thirtyone = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx".length();
System.out.println(Math.pow(two, thirtyone)-1);
How did I go? :p
我怎么去了？:p
I do like that bitshift one though...
不过我确实喜欢那个位移位...
