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?
最佳答案
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
关于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?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12634503/