位操作相当于PERL中的pop函数（删除MSB）

发布于 2024-10-08 21:00:03 字数 481 浏览 6 评论 0原文

在 Perl 中，是否有一个按位运算符，其作用类似于 >>，但删除了最高有效位？有点像 >> 运算符有点像 shift() 函数，我正在寻找一个类似于 pop()< 的位运算符/代码>。

110110 将返回 10110

101 将返回 01

最终我想看看二进制形式的数字是否是回文（即 11011、111 或 1010101），因此理想情况下，操作员有办法返回它删除的位。如果运算符不这样做也没关系，因为我可以在数学上这样做，但为了干净的代码，如果它自动返回 MSB 那就太棒了。对于LSB，我愿意

$LSB=$mynum-2*($mynum>>1);
$mynum>>=1;

原文

In Perl, is there a bitwise operator that acts like >>, but removes the most significant bit? Sort of like how the >> operator is somewhat like the shift() function, I'm looking for a bit operator that's like pop().

110110 would return 10110

101 would return 01

Ultimately I'm trying to see whether a number in binary form is palindromic (i.e. 11011, 111, or 1010101), so ideally the operator would have a way of returning the bit it removes. It's okay if the operator doesn't, as I could do so mathematically, but in the interest of clean code, it would be awesome if it returned the MSB automatically. For the LSB, I do

$LSB=$mynum-2*($mynum>>1);
$mynum>>=1;

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遇到 2024-10-15 21:00:03

我想不出比将其存储为字符串更简单的方法：

my $bits = sprintf '%b', $num;
while ( $bits =~ s/(.)// ) {
    print "removed $1\n";
}

不过你的回文检查只是

$bits eq reverse $bits

I can't think of a simpler way than just storing it as a string:

my $bits = sprintf '%b', $num;
while ( $bits =~ s/(.)// ) {
    print "removed $1\n";
}

though then your palindrome check is just

$bits eq reverse $bits

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允世 2024-10-15 21:00:03

请参阅如何检查整数的二进制表示是回文吗？

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旧瑾黎汐 2024-10-15 21:00:03

由于您的值具有可变的位数，因此您需要位字符串或位向量。查看 CPAN 上的 Bit::Vector —— 看来仍然保持活跃。

但正如其他人针对您的问题所建议的那样，处理普通的旧字符串可能会更容易。

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东走西顾 2024-10-15 21:00:03

我不了解 Perl，但在 C/C++ 中你可以这样做：

unsigned flp2(unsigned x) {
   x = x | (x >> 1);
   x = x | (x >> 2);
   x = x | (x >> 4);
   x = x | (x >> 8);
   x = x | (x >>16);
   return x - (x >> 1);
}

unsigned most_significant_bit = flp2(my_number);
my_number &= most_significant_bit;

感谢 Hacker's Delight。

请注意，要查找汇编程序中的最高有效位，您可以在 MSVC _BitScanReverse 和 GCC _builtin_clz 中使用 BSR。然而，据我所知，没有简单的高级（和便携式）运算符。语言的发展速度比 CPU 和编译器慢得多。

I don't know about Perl, but in C/C++ you'd do this like so:

unsigned flp2(unsigned x) {
   x = x | (x >> 1);
   x = x | (x >> 2);
   x = x | (x >> 4);
   x = x | (x >> 8);
   x = x | (x >>16);
   return x - (x >> 1);
}

unsigned most_significant_bit = flp2(my_number);
my_number &= most_significant_bit;

Courtesy of Hacker's Delight.

Note that to find the most significant bit in assembler you could use BSR, in MSVC _BitScanReverse, in GCC _builtin_clz. There are however no simple high-level (and portable) operators that I know of. Languages evolve much slower than CPUs and compilers.

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