Java

Question

我有一个关于 Java 中字节到 int/long 转换的查询。

有：

4 字节整数
8 字节长。

如何存储超过 8 个字节的值以及 Java 将如何表示（例如：8 + 内存中的字节值）？

Answer 1

BigInteger is designed for this. Internally, it represents numbers with a (variable-length) vector that can be seen as a polynomial. It allows arbitrary-sized integers to be manipulated efficiently, as long as there is enough free memory to do so. Other languages have similar constructs or libraries; for instance, the GMP 库遵循类似的方法。

请注意，虽然 BigIntegers 可以从 byte[] 二进制补码构建，并且可以再次转换为那些 byte[]s，但在内部它们使用更高效的 int[]表示。如有疑问，read the source.

示例代码如下：

import java.math.*;
public class T {
    public static void main(String[] args) {
        for (String s: args) {
            // build from base-16 string representation
            BigInteger bi = new BigInteger(s, 16); 
            System.out.println("The decimal representation of \n\t"
                + s + " is \n\t" + bi);
            // (bi.toString() generates base-10 strings)
        }
    }
}

现在编译 & 运行:

javac T.java && java T 8df0c57980d7fa44e3a7286cfd9a4589c5eb1eb0

预期输出：

The decimal representation of
    8df0c57980d7fa44e3a7286cfd9a4589c5eb1eb0 is 
    810337079999566280018410685437385308775358602928

Java - 将 20 字节的值转换为以 10 为基数的数字

Java - Converting 20 bytes value to base-10 number

bytearray