blocks|key|2315658|text|您可以使用逐位运算来检查gawk中的第7位是否为1|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2315659|and($2,0x40)|code-block|syntax|javascript|2315660|注意:标准awk不支持按位操作。因此，您可以使用bash位操作或perl位操作(用于字符串处理)。|2315661|使用gawk|2315662|gawk+'(and($2,0x40)){print+$1}'+filename|2315663|使用perl|2315664|perl+-ane+'print+"$F[0]\n"+if+($F[1]&0x40)'+filename|2315665|entityMap^0|C|4|O|1|0|0|5|3|O|4|W|4|0|2|4|0|0|2|4|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1D|8|@$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|V|$]]

You can use bitwise operation to check if 7th bit is <code>1</code> in <code>gawk</code>:

<pre><code>and($2,0x40)
</code></pre>

Note: Standard <code>awk</code> does not have bitwise operation. So for that you can use <code>bash</code> bitwise operation or <code>perl</code> bitwise operation (for string processing).

Using <code>gawk</code>:

<pre><code>gawk '(and($2,0x40)){print $1}' filename
</code></pre>

Using <code>perl</code>:

<pre><code>perl -ane 'print "$F[0]\n" if ($F[1]&amp;0x40)' filename
</code></pre>

blocks|key|3401634|text|bash具有按位运算符|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3401635|测试第7位：|3401636|$+echo+$(((93+&+0x40)+!=+0))
1

$+echo+$(((128+&+0x40)+!=+0))
0|code-block|syntax|javascript|3401637|另请参阅bash+documentation|offset|length|style|CODE|3401638|但是，如果要解析文件中的值，最好继续使用awk，作为@RakholiyaJenish的答案|3401639|entityMap|0|LINK|mutability|MUTABLE|url|https://www.gnu.org/software/bash/manual/bashref.html#Shell-Arithmetic^0|0|0|0|4|4|4|I|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|11|8|@$K|12|L|13|M|N]]|9|@$K|14|L|15|1|16]]|A|$]]|$1|O|3|P|5|6|7|17|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]]]

bash has bitwise operators

Test 7th bit:

<pre><code>$ echo $(((93 &amp; 0x40) != 0))
1

$ echo $(((128 &amp; 0x40) != 0))
0
</code></pre>

See also the <a href="https://www.gnu.org/software/bash/manual/bashref.html#Shell-Arithmetic" rel="noreferrer"><code>bash</code> documentation</a>

Though if you're parsing the values out of a file, you're probably better off continuing to use awk, as the answer of @RakholiyaJenish

blocks|key|3401658|text|您可以使用bash包装函数检查位集，在脚本或.bashrc中定义一个本地IFS运算符|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3401659|#+Returns+the+bit+positions+set+in+the+number+for+the+given+bit+mask
#+Use+the+return+value+of+the+function+directly+in+a+script

bitWiseAnd()+{
++++local+IFS='&'
++++printf+"%25s\n"+"$((+$*+))"
}|code-block|syntax|javascript|3401660|并在如下所示的脚本函数中使用它|3401661|#+The+arguments+to+this+function+should+be+used+as+number+and+bit-mask,+i.e.
#+bitWiseAnd+<number>+<bit-mask>
if+[+$(bitWiseAnd+"93"+"0x40")+-ne+0+]
then
++++#+other+script+actions+go+here
++++echo+"Bit+set"
fi|3401662|其思想是使用Input-Field-Separator(IFS)，这是bash中的一个特殊变量，用于在展开后拆分单词，并将行拆分为单词。该函数在本地更改该值，以使用分词字符作为逐位AND运算符&。|3401663|请记住，IFS是在本地更改的，不会对函数作用域之外的默认IFS行为产生影响。摘录自man+bash页面，|3401664|3401665|外壳程序将IFS的每个字符视为分隔符，并将其他扩展的结果拆分成这些字符上的单词。如果IFS是未设置的，或者它的值恰好是默认值，那么、和在前面的扩展结果的开头和结尾的序列将被忽略，任何不在开头或结尾的IFS字符序列用于分隔单词。|blockquote|3401666|3401667|3401668|"$((+$*+))"表示要由&拆分的参数列表，然后使用printf函数输出计算值。该函数还可以扩展为添加其他算术运算的作用域。|3401669|entityMap|0|LINK|mutability|MUTABLE|url|https://bash.cyberciti.biz/guide/$IFS^0|5|4|M|7|10|3|0|0|0|0|Z|4|2I|3|2O|1|6|Q|0|0|4|3|S|3|15|8|0|0|0|0|0|0|B|F|1|S|6|0^^$0|@$1|2|3|4|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1G|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]]|D|@$9|1P|A|1Q|1|1R]]|E|$]]|$1|Q|3|R|5|6|7|1S|8|@$9|1T|A|1U|B|C]|$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1Z|8|@]|D|@]|E|$]]|$1|T|3|U|5|V|7|20|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|21|8|@]|D|@]|E|$]]|$1|X|3|-4|5|6|7|22|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|23|8|@$9|24|A|25|B|C]|$9|26|A|27|B|C]|$9|28|A|29|B|C]]|D|@]|E|$]]|$1|10|3|-4|5|6|7|2A|8|@]|D|@]|E|$]]]|11|$12|$5|13|14|15|E|$16|17]]]]

You can use a <code>bash</code> wrapper function to check for the bit set, defining a local <code>IFS</code> operator either in a script or <code>.bashrc</code>

<pre><code># Returns the bit positions set in the number for the given bit mask
# Use the return value of the function directly in a script

bitWiseAnd() {
 local IFS='&amp;'
 printf "%s\n" "$(( $* ))"
}
</code></pre>

and use it in a function as below in your script

<pre><code># The arguments to this function should be used as number and bit-mask, i.e.
# bitWiseAnd &lt;number&gt; &lt;bit-mask&gt;
if [ $(bitWiseAnd "93" "0x40") -ne 0 ]
then
 # other script actions go here
 echo "Bit set"
fi
</code></pre>

The idea is to use the <a href="https://bash.cyberciti.biz/guide/$IFS" rel="nofollow noreferrer">Input-Field-Separator(IFS)</a>, a special variable in <code>bash</code> used for word splitting after expansion and to split lines into words. The function changes the value locally to use word-splitting character as the bit-wise <code>AND</code> operator <code>&amp;</code>.

Remember the <code>IFS</code> is changed locally and does NOT take effect on the default <code>IFS</code> behaviour outside the function scope. An excerpt from the <code>man bash</code> page,

<blockquote>
 The shell treats each character of IFS as a delimiter, and splits the results of the other expansions into words on these characters. If IFS is unset, or its value is exactly , the default, then sequences of , , and at the beginning and end of the results of the previous expansions are ignored, and any sequence of IFS characters not at the beginning or end serves to delimit words. 
</blockquote>

The <code>"$(( $* ))"</code> represents the list of arguments passed to be split by <code>&amp;</code> and later the calculated value is output using the <code>printf</code> function. The function can be extended to add scope for other arithmetic operations also.

blocks|key|2320425|text|如果作为第一个参数的数字具有作为第二个参数集的位，则此函数返回0的退出状态：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2320426|hasbitset+()+{
++++local+num=$1
++++local+bit=$2

++++if+((+num+&+2**(bit-1)+));+then
++++++++return+0
++++else
++++++++return+1
++++fi
}|code-block|syntax|javascript|2320427|或者很短，可读性较差：|2320428|hasbitset+()+{+((+$1+&+2**($2-1)+))+&&+return+0+%7C%7C+return+1;+}|2320429|对于问题中的示例：|2320430|$+hasbitset+93+7+&&+echo+"Yes"+%7C%7C+echo+"No"
Yes
$+hasbitset+128+7+&&+echo+"Yes"+%7C%7C+echo+"No"
No|2320431|请注意，通常情况下，计算偏移量中的位数而不是位置中的位数，即从0位开始-不同于此问题中的用法。|2320432|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

This function returns an exit status of zero if the number given as its first argument has the bit given as its second argument set:

<pre><code>hasbitset () {
 local num=$1
 local bit=$2

 if (( num &amp; 2**(bit-1) )); then
 return 0
 else
 return 1
 fi
}
</code></pre>

or short and less readable:

<pre><code>hasbitset () { (( $1 &amp; 2**($2-1) )) &amp;&amp; return 0 || return 1; }
</code></pre>

For the examples from the question:

<pre><code>$ hasbitset 93 7 &amp;&amp; echo "Yes" || echo "No"
Yes
$ hasbitset 128 7 &amp;&amp; echo "Yes" || echo "No"
No
</code></pre>

Notice that it's often customary to count bits in offsets instead of positions, i.e., starting from bit 0 &ndash; unlike the usage in this question.

I'm wondering if there's a way to replace the if statement with something that checks whether <code>$2</code> has the 7th bit set to 1?

<pre><code>cat $file | awk '{if ($2 == 87) print $1; else {}}' &gt; out.txt"
</code></pre>

For instance, 93 should print something whereas 128 should not.

Check if nth bit is set in bash

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想知道是否有一种方法可以用来检查$2是否将第7位设置为1来替换if语句？cat $file | awk '{if ($2 == 87) print $1; else {}}' > out.txt"例如，93应该打印一些东西，而128不应该。

问检查bash中是否设置了第n位
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问检查bash中是否设置了第n位EN