bash - 为什么对两个大的二进制数进行异或运算会得到十进制输出？

Question

我正在尝试对两个大的二进制值进行异或运算。然而，

echo ${hashArray[1]}
echo ${hashArray[2]}
echo $((${hashArray[1]} ^ ${hashArray[2]}))

给出:

10100100000111101011100001101110000110000100001000000111001001100010110000010010111101100110111001111100010010000000010101110111

00001110110000010110101101011011100101101000011100011101001101101010000100110001001110101101111100010001111010100011010000000100

4018181242766406943

为什么 echo $((${hashArray[1]} ^ ${hashArray[2]})) 输出一个十进制数？它不应该是另一个大的二进制值吗？

Answer 1

bash 确实有 support for binary numbers ，但是您的数字太大，将被截断。要么按照 mana 的建议去做，要么拆分字符串，要么使用可以处理任意精度的库，例如想到 perl 的 Math::BigInt:

xor.pl

use Math::BigInt;

$x = Math::BigInt->new("0b" . $ARGV[0]); 
$y = Math::BigInt->new("0b" . $ARGV[1]);

print $x->bxor($y)->as_bin;

为方便起见，将数字分配给变量:

x=10100100000111101011100001101110000110000100001000000111001001100010110000010010111101100110111001111100010010000000010101110111
y=00001110110000010110101101011011100101101000011100011101001101101010000100110001001110101101111100010001111010100011010000000100

运行:

perl xor.pl $x $y

输出:

0b10101010110111111101001100110101100011101100010100011010000100001000110100100011110011001011000101101101101000100011000101110011