我无法将 Hex 转换为 BigInteger

Question

我尝试将 Hex 转换为 BigInteger，这里是源代码：

public static void main (String[] args)
{
    BigInteger a = new BigInteger("c6e87a3b3ef791287ac85a0c7836dd8305dd9a4c0024123db9145bb5067a8abf142e9001b0788039fd4d676b59db2110da23532282c7648d94fdbf29b731b0d21f9ca51acd44063f271326915283af97f0822519bbe2a6b80618e45e6194b2445d5afe70cf2c10569034966f3bc3b9a30d3ac4f06dbbca89fce7ef64ee14de3dL", 16);
    BigInteger b = new BigInteger("9e5d1bd4f53eebc8a695c61ba4436e38af273fd6733115611fded8dd407b5f0bc04301829dc6ed921af866c3c7977839fc75831152307f8e50e3c0f9107b6ae82ddab584807ea5ba7f32f9bfcab6218c6c6367817dfdd3b2ccc5c21cc9550b9248cac34dfb0d22151c196ca843f15614b3f6b044f9c5e727dc0b44f441c2ed7fL", 16);
    System.out.println(a);
    System.out.println(b);
    // System.out.println(modInv(a, b));
}

我尝试运行这个源代码，但是我得到了这样的运行时错误：

Exception in thread "main" java.lang.NumberFormatException: For input string: "....14de3dL" at java.lang.NumberFormatException.forInputString(Unknown Source) at java.lang.Integer.parseInt(Unknown Source) at java.math.BigInteger.(Unknown Source) at sss.main.main(main.java:15)

源代码有问题吗？我找不到。

Answer 1

首先，L 不包含在您写在两个字符串末尾的 Hex Numbers 中。您可以访问此 link 以获得有关 Hex Numbers

的帮助

https://en.wikipedia.org/wiki/Hexadecimal

访问后，请同时查看此post

Java convert a HEX String to a BigInt

Answer 2

正如注释中指出的，L 不是十六进制的一部分 presentation.L 用于将字符串表示为 Long，因此在十六进制表示中它是无效字符。下面的代码应该会有帮助

 BigInteger a = new BigInteger("c6e87a3b3ef791287ac85a0c7836dd8305dd9a4c0024123db9145bb5067a8abf142e9001b0788039fd4d676b59db2110da23532282c7648d94fdbf29b731b0d21f9ca51acd44063f271326915283af97f0822519bbe2a6b80618e45e6194b2445d5afe70cf2c10569034966f3bc3b9a30d3ac4f06dbbca89fce7ef64ee14de3d", 16);
 System.out.println(a.toString());

我无法将 Hex 转换为 BigInteger

I can't convert Hex to BigInteger

java

hex

biginteger