blocks|key|645830|text|JavaScript，28个字符|type|header-two|depth|inlineStyleRanges|entityRanges|data|645831|for(y=1;y%2B1-1;y/=2);alert(y)|code-block|syntax|javascript|645832|我可以用y%2B1-1代替y%2B1>1，但我喜欢前者的美学。|unstyled|offset|length|style|CODE|entityMap^0|0|0|4|5|B|5^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|Q|8|@$J|R|K|S|L|M]|$J|T|K|U|L|M]]|9|@]|A|$]]]|N|$]]

<h2>JavaScript, 28 characters</h2>

<pre class="lang-js prettyprint-override"><code>for(y=1;y+1-1;y/=2);alert(y)
</code></pre>

<p>I could replace the <code>y+1-1</code> with <code>y+1&gt;1</code>, but I like for aesthetics of the former.</p>


blocks|key|642345|text|Brainfuck，16岁，|type|header-two|depth|inlineStyleRanges|entityRanges|data|642346|Brainfuck不支持浮点运算，所以epsilon是0.|unstyled|642347|-[>-<%2B%2B%2B%2B%2B]>---.|code-block|syntax|javascript|642348|根据这的说法，这是最小的这类程序。程序假定单元格是无符号字节mod+256。|offset|length|entityMap|0|LINK|mutability|MUTABLE|url|http://www.esolangs.org/wiki/Brainfuck_constants^0|0|0|0|2|1|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|W|8|@]|9|@]|A|$H|I]]|$1|J|3|K|5|D|7|X|8|@]|9|@$L|Y|M|Z|1|10]]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<p><strong>Brainfuck, 16</strong></p>

<p>Brainfuck doesn't support floating-point arithmetic, so epsilon is 0.</p>

<pre><code>-[&gt;-&lt;+++++]&gt;---.
</code></pre>

<p>According to <a href="http://www.esolangs.org/wiki/Brainfuck_constants" rel="nofollow">this</a>, this is the smallest such program.  Program assumes that cells are unsigned bytes mod 256.</p>


blocks|key|645878|text|Python，29|type|header-two|depth|inlineStyleRanges|entityRanges|data|645879|t=1.
while+t%2B1>1:t/=2
print+t|code-block|syntax|javascript|645880|Python3，28|645881|“真”组织糟透了，海事组织.但我会用它来删掉一个角色|unstyled|645882|t=1
while+t%2B1>1:t/=2
print+t|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|R|8|@]|9|@]|A|$]]|$1|L|3|M|5|D|7|S|8|@]|9|@]|A|$E|F]]]|N|$]]

<p><strong>Python, 29</strong></p>

<pre><code>t=1.
while t+1&gt;1:t/=2
print t
</code></pre>

<p><strong>Python3, 28</strong></p>

<p>"true" division sucks, IMO... but I'll use it to cut a character out</p>

<pre><code>t=1
while t+1&gt;1:t/=2
print t
</code></pre>


blocks|key|646015|text|PHP，13个字符|type|header-two|depth|inlineStyleRanges|entityRanges|data|646016|要使1+%2B+epsilon+>+1为真，1+%2B+epsilon必须符合以浮点数分配给尾数(5+in+5+*+10%5E4)的位数。以64位浮点数分配给尾数52位.因此，1+%2B+epsilon是1.000...(52)...0001+(基2)，episilon是0.000...(52)...0001。用十进制计算这个数字可以得到：|unstyled|offset|length|style|CODE|646017|<?=pow(2,-52);|code-block|syntax|javascript|646018|你会注意到，负指数2和重复除以2是一样的--这就是这里很多解的作用。|646019|奖金，更独立于平台的解决方案(13个字符)：|646020|<?=sin(pi());|entityMap^0|0|2|F|K|B|19|1|1E|8|2A|B|2M|J|3B|8|3K|J|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@$E|W|F|X|G|H]|$E|Y|F|Z|G|H]|$E|10|F|11|G|H]|$E|12|F|13|G|H]|$E|14|F|15|G|H]|$E|16|F|17|G|H]|$E|18|F|19|G|H]|$E|1A|F|1B|G|H]]|9|@]|A|$]]|$1|I|3|J|5|K|7|1C|8|@]|9|@]|A|$L|M]]|$1|N|3|O|5|D|7|1D|8|@]|9|@]|A|$]]|$1|P|3|Q|5|D|7|1E|8|@]|9|@]|A|$]]|$1|R|3|S|5|K|7|1F|8|@]|9|@]|A|$L|M]]]|T|$]]

<h2>PHP, 13 chars</h2>

<p>For <code>1 + epsilon &gt; 1</code>, to be true, <code>1 + epsilon</code> must fit into the number of bits allotted to the mantissa (<code>5</code> in <code>5 * 10^4</code>) in floating point numbers. 52 bits are allotted to the mantissa in 64-bit floating point numbers. It follows that <code>1 + epsilon</code> is <code>1.000...(52)...0001</code> (base 2), and that <code>episilon</code> is <code>0.000...(52)...0001</code>. Calculating this number in decimal gives:</p>

<pre><code>&lt;?=pow(2,-52);
</code></pre>

<p>You'll notice that negatively exponentiating 2 is the same as repeatedly dividing by 2 -- which is what many of the solutions here do.</p>

<p>Bonus, more platform-independent solution (13 chars):</p>

<pre><code>&lt;?=sin(pi());
</code></pre>


blocks|key|642709|text|MATLAB，3字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|642710|我将发布这纯粹是因为这是一个难得的机会，MATLAB成为其中之一，如果不是最短的。我认为这是问题的规则。|unstyled|642711|eps|code-block|syntax|javascript|642712|当然，并不是最巧妙或最令人兴奋的代码！|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|O|8|@]|9|@]|A|$H|I]]|$1|J|3|K|5|D|7|P|8|@]|9|@]|A|$]]]|L|$]]

<h1>MATLAB, 3 bytes</h1>
<p>I'm going to post this purely because it's one of the rare opportunities for MATLAB to be one of if not the shortest. I think it is in the rules for what the question asks.</p>
<pre><code>eps
</code></pre>
<p>Granted, not the most ingenious or exciting code though!</p>


blocks|key|642429|text|Mathematica+15+chars|type|header-two|depth|inlineStyleRanges|entityRanges|data|642430|$MachineEpsilon

->+2.220446049250313*10%5E-16|code-block|syntax|javascript|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|I|8|@]|9|@]|A|$E|F]]]|G|$]]

<h2>Mathematica 15 chars</h2>

<pre><code>$MachineEpsilon

-&gt; 2.220446049250313*10^-16
</code></pre>


blocks|key|642584|text|Ada+-+151|type|header-two|depth|inlineStyleRanges|entityRanges|data|642585|generic
type+T+is+digits+<>;function+E+return+T;function+E+return+T+is
A,B,C:T:=1.0;begin
loop
A:=A/2.0;B:=C%2BA;exit+when+B<=C;end+loop;return+A;end+E;|code-block|syntax|javascript|642586|示例用法：|unstyled|642587|procedure+Main+is
+++function+F+is+new+E+(Float);
+++function+D+is+new+E+(Long_Float);
begin
+++Put_Line+("Epsilon:"+&+Float'Image+(F));
+++Put_Line+("Epsilon:"+&+Long_Float'Image+(D));
end+Main;|642588|产出：|642589|Epsilon:+5.96046E-08
Epsilon:+1.11022302462516E-16|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|S|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|L|3|M|5|I|7|U|8|@]|9|@]|A|$]]|$1|N|3|O|5|D|7|V|8|@]|9|@]|A|$E|F]]]|P|$]]

<h2>Ada - 151</h2>

<pre><code>generic
type T is digits &lt;&gt;;function E return T;function E return T is
A,B,C:T:=1.0;begin
loop
A:=A/2.0;B:=C+A;exit when B&lt;=C;end loop;return A;end E;
</code></pre>

<p>example usage:</p>

<pre><code>procedure Main is
   function F is new E (Float);
   function D is new E (Long_Float);
begin
   Put_Line ("Epsilon:" &amp; Float'Image (F));
   Put_Line ("Epsilon:" &amp; Long_Float'Image (D));
end Main;
</code></pre>

<p>output:</p>

<pre><code>Epsilon: 5.96046E-08
Epsilon: 1.11022302462516E-16
</code></pre>


blocks|key|646103|text|Pyth+-5字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|646104|.IhG1|code-block|syntax|javascript|646105|在网上试一试。|unstyled|offset|length|646106|倒置通过所有的正反应，直到它找到一些东西，使G的λ等于第二arg。因此，代码只是反转1%2BG，直到它等于1。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|http://pyth.herokuapp.com/?code=.IhG1&debug=0^0|0|0|0|6|0|0|M|1|16|3|1F|1^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|Y|8|@]|9|@$J|Z|K|10|1|11]]|A|$]]|$1|L|3|M|5|I|7|12|8|@$J|13|K|14|N|O]|$J|15|K|16|N|O]|$J|17|K|18|N|O]]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

<h1>Pyth - 5 bytes</h1>
<pre><code>.IhG1
</code></pre>
<p><a href="http://pyth.herokuapp.com/?code=.IhG1&amp;debug=0" rel="nofollow noreferrer">Try it online here</a>.</p>
<p>Invert goes through all positive reals till it finds something that makes the lambda of <code>G</code> equal the second arg. So the code just inverts <code>1+G</code> till it equals <code>1</code>.</p>


blocks|key|646217|text|Excel，35字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|646218|=LET(x,2%5E-ROW(1:99),MAX((1%2Bx=1)*x))|code-block|syntax|javascript|646219|Excel+(2010)，37字节|646220|=MAX(2%5E-ROW(1:99)*(1%2B2%5E-ROW(1:99)=1))|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|M|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|N|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|O|8|@]|9|@]|A|$E|F]]]|K|$]]

<h1>Excel, 35 bytes</h1>
<pre><code>=LET(x,2^-ROW(1:99),MAX((1+x=1)*x))
</code></pre>
<h2>Excel (2010), 37 bytes</h2>
<pre><code>=MAX(2^-ROW(1:99)*(1+2^-ROW(1:99)=1))
</code></pre>


blocks|key|646246|text|Java，13字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|646247|使用ulp方法很容易。|unstyled|offset|length|646248|Math.ulp(1.0)|code-block|syntax|javascript|646249|结果为2.220446049250313E-16。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html#ulp-double-^0|0|2|3|0|0|0|3|L^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@$E|Y|F|Z|1|10]]|A|$]]|$1|G|3|H|5|I|7|11|8|@]|9|@]|A|$J|K]]|$1|L|3|M|5|D|7|12|8|@$E|13|F|14|N|O]]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

<h1>Java, 13 bytes</h1>
<p>It is easy by using the <a href="https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html#ulp-double-" rel="nofollow noreferrer">ulp</a> method.</p>
<pre><code>Math.ulp(1.0)
</code></pre>
<p>Result is <code>2.220446049250313E-16</code>.</p>


blocks|key|646622|text|外壳，8字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|646623|Ωoε→½□√2|code-block|syntax|javascript|646624|在网上试试！|unstyled|646625|公式与参考实现完全相同。|646626|用小数修正了利奥的帮助。|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/barbuz/Husk|1|https://tio.run/##ARkA5v9odXNr///OqW/OteKGksK94pah4oiaMv//^0|0|2|0|0|0|0|6|1|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@$A|14|B|15|1|16]]|C|$]]|$1|L|3|M|5|K|7|17|8|@]|9|@]|C|$]]|$1|N|3|O|5|K|7|18|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h1><a href="https://github.com/barbuz/Husk" rel="nofollow noreferrer">Husk</a>, 8 bytes</h1>
<pre><code>Ωoε→½□√2
</code></pre>
<p><a href="https://tio.run/##ARkA5v9odXNr///OqW/OteKGksK94pah4oiaMv//" rel="nofollow noreferrer" title="Husk – Try It Online">Try it online!</a></p>
<p>Formula is the exact same as the reference implementation.</p>
<p>Fixed with Leo's help by using decimals.</p>


blocks|key|646740|text|格罗克，21字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|646741|1Y++j+qzph
>1%2B1hY/2p{|code-block|syntax|javascript|646742|在网上试试！|unstyled|646743|是的，现在有了一个在线翻译！对于那些好奇的人来说，这个框架是从在线Vyxal解释器中修改的。|646744|解释：|646745|1Y++j+++++++#+Push+1+and+copy+it+to+the+register
++%2B1h+++++++#+Add+1
>1++++++++++#+Is+it+greater+than+1?
++++++++p{++#+If+yes,+retrieve+value+from+register
+++++Y/2++++#+Divide+by+2+and+copy+it+to+the+register,+then+repeat+this+line
++++++qzph++#+If+no,+retrieve+value+from+register,
+++++++++{++#+then+print+as+a+number|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/AMiller42/Grok-Language|1|http://grok.pythonanywhere.com/?flags=&code=1Y%2520%2520j%2520qzph%250A%253E1%252B1hY%252F2p%257B&inputs=^0|0|3|0|0|0|0|6|1|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|L|3|M|5|K|7|19|8|@]|9|@]|C|$]]|$1|N|3|O|5|K|7|1A|8|@]|9|@]|C|$]]|$1|P|3|Q|5|F|7|1B|8|@]|9|@]|C|$G|H]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]]]

<h1><a href="https://github.com/AMiller42/Grok-Language" rel="nofollow noreferrer">Grok</a>, 21 bytes</h1>
<pre><code>1Y  j qzph
&gt;1+1hY/2p{
</code></pre>
<p><a href="http://grok.pythonanywhere.com?flags=&amp;code=1Y%20%20j%20qzph%0A%3E1%2B1hY%2F2p%7B&amp;inputs=" rel="nofollow noreferrer">Try it Online!</a></p>
<p>That's right, there's an online interpreter now! For those wondering, the framework is modified from the online Vyxal interpreter.</p>
<p>Explanation:</p>
<pre><code>1Y  j       # Push 1 and copy it to the register
  +1h       # Add 1
&gt;1          # Is it greater than 1?
        p{  # If yes, retrieve value from register
     Y/2    # Divide by 2 and copy it to the register, then repeat this line
      qzph  # If no, retrieve value from register,
         {  # then print as a number
</code></pre>


blocks|key|642654|text|Ruby，18岁，|type|header-two|depth|inlineStyleRanges|entityRanges|data|642655|只是为了完整：|unstyled|642656|p+Float::EPSILON/2|code-block|syntax|javascript|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|M|8|@]|9|@]|A|$H|I]]]|J|$]]

<h2>Ruby, 18</h2>

<p>Just for the sake of completeness:</p>

<pre><code>p Float::EPSILON/2
</code></pre>


blocks|key|646038|text|Mathematica，21字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|646039|Print@$MachineEpsilon|code-block|syntax|javascript|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|I|8|@]|9|@]|A|$E|F]]]|G|$]]

<h1>Mathematica, 21 bytes</h1>

<pre><code>Print@$MachineEpsilon
</code></pre>


blocks|key|642774|text|JavaScript+ES6+in+a+REPL,14|type|header-two|depth|inlineStyleRanges|entityRanges|data|642775|Number.EPSILON|code-block|syntax|javascript|642776|REPL中的|unstyled|642777|JavaScript+ES5，15|642778|Math.pow(2,-52)|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|Q|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|R|8|@]|9|@]|A|$]]|$1|L|3|M|5|D|7|S|8|@]|9|@]|A|$E|F]]]|N|$]]

<h1>JavaScript ES6 in a REPL,14</h1>

<pre><code>Number.EPSILON
</code></pre>

<h1>JavaScript ES5 in a REPL,15</h1>

<pre><code>Math.pow(2,-52)
</code></pre>


blocks|key|642855|text|C%2B%2B+(gcc)，63字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|642856|#import<iostream>
int+main(){long+x=1;std::cout<<*(double*)&x;}|code-block|syntax|javascript|642857|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://gcc.gnu.org/|1|https://tio.run/##Sy4o0E1PTv7/XzkztyC/qMQmM7@4pCg1MdeOKzOvRCE3MTNPQ7M6Jz8vXaHC1tC6uCTFyio5v7TExkZLIyW/NCknVUtTrcK69v9/AA^0|0|9|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="https://gcc.gnu.org/" rel="nofollow noreferrer">C++ (gcc)</a>, 63 bytes</h1>

<pre class="lang-cpp prettyprint-override"><code>#import&lt;iostream&gt;
int main(){long x=1;std::cout&lt;&lt;*(double*)&amp;x;}
</code></pre>
<p><a href="https://tio.run/##Sy4o0E1PTv7/XzkztyC/qMQmM7@4pCg1MdeOKzOvRCE3MTNPQ7M6Jz8vXaHC1tC6uCTFyio5v7TExkZLIyW/NCknVUtTrcK69v9/AA" rel="nofollow noreferrer" title="C++ (gcc) – Try It Online">Try it online!</a></p>


blocks|key|646374|text|GNU+AWK，34字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|646375|BEGIN{for(e=1;1%2Be>1;e/=2);print+e}|code-block|syntax|javascript|646376|在网上试试！不同精度的感应器具有不同的特性。|unstyled|646377|由于GNU算法，GAWK可以依赖任何任意字节的精度。GAWK的手册参考.标准精度(双)返回1.11022e-16。|style|CODE|646378|Result:
0.000488281+half+precision
5.96046e-08+single+precision
1.11022e-16+double+precision
9.62965e-35+quad+precision
4.52784e-72+oct+precision|entityMap|0|LINK|mutability|MUTABLE|url|https://www.gnu.org/software/gawk/manual/gawk.html|1|https://tio.run/##pZHNSsNAFIX38xSXdNNipyXqLqRQpRYXU0rBB5hM7iRDJzNxfigifXXHBBfqRpBs7/nOx4HLL@eUHnb758O7tG6OZV7kN7jJC1yXt4uid8oEwCvMgL3B3T0VLXdcBHTQWC3BWx2DsoYQMoOtqaFWUqLDsdR7GJQ/Lr1DofxA@xUwNcChRaAM5pRWqjGxWwB3TexGNlhAwyuNsD@8ADs@nYZMhbbDoMSKkK/Jx9PuscxarmVW/DF/CSOYfQ/Irr8EXplG4yRFbWM1UfEaeT1JYEX4bz@lD9uPD/SJsk8|2|https://www.gnu.org/software/gawk/manual/gawk.html#Setting-precision^0|0|7|0|0|0|0|M|1|0|19|B|Q|A|2|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]|$1|L|3|M|5|K|7|1B|8|@$A|1C|B|1D|N|O]]|9|@$A|1E|B|1F|1|1G]]|C|$]]|$1|P|3|Q|5|F|7|1H|8|@]|9|@]|C|$G|H]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]|10|$5|T|U|V|C|$W|11]]]]

<h1><a href="https://www.gnu.org/software/gawk/manual/gawk.html" rel="nofollow noreferrer">GNU AWK</a>, 34 bytes</h1>
<pre><code>BEGIN{for(e=1;1+e&gt;1;e/=2);print e}
</code></pre>
<p><a href="https://tio.run/##pZHNSsNAFIX38xSXdNNipyXqLqRQpRYXU0rBB5hM7iRDJzNxfigifXXHBBfqRpBs7/nOx4HLL@eUHnb758O7tG6OZV7kN7jJC1yXt4uid8oEwCvMgL3B3T0VLXdcBHTQWC3BWx2DsoYQMoOtqaFWUqLDsdR7GJQ/Lr1DofxA@xUwNcChRaAM5pRWqjGxWwB3TexGNlhAwyuNsD@8ADs@nYZMhbbDoMSKkK/Jx9PuscxarmVW/DF/CSOYfQ/Irr8EXplG4yRFbWM1UfEaeT1JYEX4bz@lD9uPD/SJsk8" rel="nofollow noreferrer" title="AWK – Try It Online">Try it online! Features different epsilons for different precisions.</a></p>
<p>Thanks to GNU MPFR arithmetic, GAWK can rely on any arbitrary byte precision. <a href="https://www.gnu.org/software/gawk/manual/gawk.html#Setting-precision" rel="nofollow noreferrer">GAWK's manual reference.</a>
The standard precision (double) returns <code>1.11022e-16</code>.</p>
<pre><code>Result:
0.000488281 half precision
5.96046e-08 single precision
1.11022e-16 double precision
9.62965e-35 quad precision
4.52784e-72 oct precision
</code></pre>


blocks|key|646490|text|05AB1E，9字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|646491|1[D>1›_#;|code-block|syntax|javascript|646492|在网上试试！|unstyled|646493|1[D>1›_#;++#+full+program
+[++++_#+++#+while...
1+D++++++++#+literal...
++D++++++++#+or+top+of+stack+if+not+first+iteration...
+++>+++++++#+plus+1...
+++++›+++++#+is+greater+than...
++++1++++++#+literal...
++++++++;++#+halve+top+of+stack
+++++++++++#+implicit+output|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Adriandmen/05AB1E|1|https://tio.run/##yy9OTMpM/f/fMNrFzvBRw654Zev//wE^0|0|6|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|F|7|15|8|@]|9|@]|C|$G|H]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1><a href="https://github.com/Adriandmen/05AB1E" rel="nofollow noreferrer">05AB1E</a>, 9 bytes</h1>
<pre><code>1[D&gt;1›_#;
</code></pre>
<p><a href="https://tio.run/##yy9OTMpM/f/fMNrFzvBRw654Zev//wE" rel="nofollow noreferrer" title="05AB1E – Try It Online">Try it online!</a></p>
<pre><code>1[D&gt;1›_#;  # full program
 [    _#   # while...
1 D        # literal...
  D        # or top of stack if not first iteration...
   &gt;       # plus 1...
     ›     # is greater than...
    1      # literal...
        ;  # halve top of stack
           # implicit output
</code></pre>


blocks|key|643209|text|JavaScript+(ES6)，24字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|643210|(e=n=>n%2B1>1?e(n/2):n)(1)|code-block|syntax|javascript|643211|在网上试试！|unstyled|offset|length|643212|递归解决方案，使用与问题相同的算法。|643213|它的身体是一个生命，所以它是一个表达式，它直接计算出epsilon值。|643214|(
++e+=+n+=>+++++++//Set+e+to+the+function,+so+we+can+call+it+later
++++n+%2B+1+>+1++++//If+n%2B1+is+still+greater+than+1...
++++++?+e(n+/+2)+//...perform+next+iteration+by+calling+itself+with+n+halved
++++++:+n++++++++//...else,+we+found+epsilon,+return+it
)(1)+++++++++++++//Call+the+function+with+the+initial+value+1|643215|有趣的事实:这将返回Number.EPSILON的一半，因为它的值代表可以表示的最小数目，即Number.EPSILON+%2B+1+>+1，而Number.EPSILON/2+%2B+1+===+1。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://tio.run/##y0osSyxOLsosKNHNy09J/Z@cn1ecn5Oql5OfrvFfI9U2z9YuT9vQztA@VSNP30jTKk9Tw1Dzv@Z/AA^0|0|0|0|6|0|0|0|0|0|A|E|1A|M|1Y|Q^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|14|8|@]|9|@$J|15|K|16|1|17]]|A|$]]|$1|L|3|M|5|I|7|18|8|@]|9|@]|A|$]]|$1|N|3|O|5|I|7|19|8|@]|9|@]|A|$]]|$1|P|3|Q|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|R|3|S|5|I|7|1B|8|@$J|1C|K|1D|T|U]|$J|1E|K|1F|T|U]|$J|1G|K|1H|T|U]]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]]]

<h2>JavaScript (ES6), 24 bytes</h2>
<pre><code>(e=n=&gt;n+1&gt;1?e(n/2):n)(1)
</code></pre>
<p><a href="https://tio.run/##y0osSyxOLsosKNHNy09J/Z@cn1ecn5Oql5OfrvFfI9U2z9YuT9vQztA@VSNP30jTKk9Tw1Dzv@Z/AA" rel="nofollow noreferrer" title="JavaScript (Node.js) – Try It Online">Try it online!</a></p>
<p>Recursive solution, using the same algorithm as in the question.</p>
<p>Its body is an IIFE, so it's an expression that directly evaluates to the epsilon value.</p>
<pre class="lang-js prettyprint-override"><code>(
  e = n =&gt;       //Set e to the function, so we can call it later
    n + 1 &gt; 1    //If n+1 is still greater than 1...
      ? e(n / 2) //...perform next iteration by calling itself with n halved
      : n        //...else, we found epsilon, return it
)(1)             //Call the function with the initial value 1
</code></pre>
<p>Fun fact: this will return the <em>half</em> of <code>Number.EPSILON</code>, as its value represents the smallest number that <em>can</em> be represented, i.e. <code>Number.EPSILON + 1 &gt; 1</code>, while <code>Number.EPSILON/2 + 1 === 1</code>.</p>


blocks|key|643262|text|PostScript，37字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|643263|([){2+div+dup+1+add+1+gt{[}if}def+1[=|code-block|syntax|javascript|643264|运行鬼怪脚本(使用可选的-q来清理输出)：|unstyled|643265|$+gs+-q+epsilon.ps+
5.96046e-08
GS>|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|O|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|P|8|@]|9|@]|A|$E|F]]]|L|$]]

<h1>PostScript, 37 bytes</h1>
<pre><code>([){2 div dup 1 add 1 gt{[}if}def 1[=
</code></pre>
<p>Run through ghostscript (with optional -q to clean up output):</p>
<pre><code>$ gs -q epsilon.ps 
5.96046e-08
GS&gt;
</code></pre>


<p><a href="http://en.wikipedia.org/wiki/Machine_epsilon" rel="nofollow">Machine epsilon</a> is an important floating point number to know when doing numerical calculations. One way to understand it is when this relation</p>

<pre><code>1 + machine_epsilon &gt; 1
</code></pre>

<p>does not hold. One (iterative) way to extract it is executing a small program like the one above:</p>

<pre><code>10 EPS=1
20 EPS=EPS/2
30 EPS1 = 1+EPS
40 IF EPS1 &gt; 1 GOTO 20
50 PRINT EPS
</code></pre>

<p>But there may be more ways. Write the shortest code to extract the machine epsilon of your computer. <strong>NOTE:</strong> epsilon should be a non-zero value :)</p>


Machine Epsilon

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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在进行数值计算时，机器感应器是一个重要的浮点数字。一种理解这种关系的方法是1 + machine_epsilon > 1不支持。提取它的一种(迭代)方法是执行像上面这样的小程序：10 EPS=120 EPS=EPS/230 EPS1 = 1+EPS40 IF EPS1 > 1 GOTO 2050 PRINT EPS但也许还有更多的方法。编写最短的代码来提取计算机的机器感应器。注: epsilon应

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