为什么Java能够将0xff000000存储为int？
Java 中整数的最大值是 2147483647，因为 Java 整数是有符号的，对吧？
0xff000000 的数值为 4278190080。
但是我看到这样的 Java 代码：
int ALPHA_MASK = 0xff000000;
谁能启发我？
An integer's max value in Java is 2147483647, since Java integers are signed, right?
0xff000000 has a numeric value of 4278190080.
Yet I see Java code like this:
int ALPHA_MASK = 0xff000000;
Can anyone enlighten me please?
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只是对埃里克森的答案的补充：
正如他所说，在大多数计算机体系结构上，有符号整数存储为各自正值的补码。
也就是说，全部 2^32 个可能的值被分为两组：一组代表从 0 位开始的正值，一组代表从 1 开始的负值。
现在，想象一下我们仅限于 3 位数字。 让我们以一种有趣的方式排列它们，很快就会有意义：
您会看到左侧的所有数字都以 1 位开头，而右侧的所有数字都以 0 开头。根据我们之前的决定为了将前者声明为负数，将后者声明为正数，我们看到 001、010 和 011 是唯一可能的正数，而 111、110 和 101 则是它们各自的负数。
现在我们分别如何处理顶部和底部的两个数字？ 显然，000 应该为零，而 100 将是所有没有正数对应项的最小负数。 总结一下：
您可能会注意到，可以通过对 1 (001) 求反并添加 1 (001) 来获得 -1 (111) 的位模式：
001（= 1）-> 110+001-> 111 (= -1)
回到你的问题：
0xff000000 = 1111 1111 0000 0000 0000 0000 0000 0000
我们不必在它前面添加更多的零，因为我们已经达到了 32 位的最大值。
另外，它显然是一个负数（因为它以 1 位开始），所以我们现在要计算它的绝对值/正对应值：
这意味着，我们将取其二进制补码，
然后
我们添加
和因此
，0xff000000 = -16777216。
Just an addition to erickson's answer:
As he said, signed integers are stored as two's complements to their respective positive value on most computer architectures.
That is, the whole 2^32 possible values are split up into two sets: one for positive values starting with a 0-bit and one for negative values starting with a 1.
Now, imagine that we're limited to 3-bit numbers. Let's arrange them in a funny way that'll make sense in a second:
You see that all numbers on the left-hand side start with a 1-bit whereas on the right-hand side they start with a 0. By our earlier decision to declare the former as negative and the latter as positive, we see that 001, 010 and 011 are the only possible positive numbers whereas 111, 110 and 101 are their respective negative counterparts.
Now what do we do with the two numbers that are at the top and the bottom, respectively? 000 should be zero, obviously, and 100 will be the lowest negative number of all which doesn't have a positive counterpart. To summarize:
You might notice that you can get the bit pattern of -1 (111) by negating 1 (001) and adding 1 (001) to it:
001 (= 1) -> 110 + 001 -> 111 (= -1)
Coming back to your question:
0xff000000 = 1111 1111 0000 0000 0000 0000 0000 0000
We don't have to add further zeros in front of it as we already reached the maximum of 32 bits.
Also, it's obviously a negative number (as it's starting with a 1-bit), so we're now going to calculate its absolute value / positive counterpart:
This means, we'll take the two's complement of
which is
Then we add
and obtain
Therefore, 0xff000000 = -16777216.
高位是符号位。 设置为负数：-16777216。
与大多数语言一样，Java 以 2 的补码 形式存储带符号的数字。 在本例中，从 0x7F000000（即 2130706432）减去 231（即 2147483648）将得到 -16777216。
The high bit is a sign bit. Setting it denotes a negative number: -16777216.
Java, like most languages, stores signed numbers in 2's complement form. In this case, subtracting 231, or 2147483648 from 0x7F000000, or 2130706432, yields -16777216.
可能值得指出的事情 - 该代码并不意味着用作具有数值的整数； 目的是作为位掩码从 32 位颜色值中过滤掉 Alpha 通道。 这个变量实际上甚至不应该被视为一个数字，而只是一个高 8 位打开的二进制掩码。
Something probably worth pointing out - this code is not meant to be used as an integer with a numerical value; The purpose is as a bitmask to filter the alpha channel out of a 32 bit color value. This variable really shouldn't even be thought of as a number, just as a binary mask with the high 8 bits turned on.
额外的位用于符号
Java ints are 二进制补码
the extra bit is for the sign
Java ints are twos complement
Java 中的 int 是有符号的。
ints are signed in Java.