entityMap|blocks|key|6irlk|text|您正在使用long调用Math.sqrt。正如JavaDoc所指出的，它返回一个“正确四舍五入的值”。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|bpiph|由于平方根不是非逗号(87522238,999999994)，因此将结果向上舍入到输出87522239。|coh8i|在此之后，该值的平方直观地大于mynum，因为您要乘以比根更大的数字！|a4sm4^0|B|9|0|B|I|17|8|0|F|5|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@$A|M|B|N|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|O|9|@$A|P|B|Q|C|D]|$A|R|B|S|C|D]]|E|@]|F|$]]|$2|I|4|J|6|7|8|T|9|@$A|U|B|V|C|D]]|E|@]|F|$]]|$2|K|4|-4|6|7|8|W|9|@]|E|@]|F|$]]]]

You are calling <code>Math.sqrt</code> with a long.
As the JavaDoc points out, it returns a "correctly rounded value". 

Since your square-root is not an non-comma-value (<code>87522238,999999994</code>), your result is rounded up to your output <code>87522239</code>.

After that, the square of the value is intuitively larger than <code>mynum</code>, since you multiply larger numbers than the root!

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://en.wikipedia.org/wiki/Double-precision_floating-point_format|1|https://en.wikipedia.org/wiki/Integer_square_root|2|https://github.com/google/guava/blob/master/guava/src/com/google/common/math/LongMath.java#L312|blocks|key|5fino|text|Math.sqrt在double%25s上运行，而不是在long%25s上运行，因此首先将mynum转换为double。这是一个64位的浮点数，它有“15-17个十进制数字的精度”(Wikipedia)。|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|8nkk7|您的输入数字为16位，因此您可能已经失去了输入的精度。您可能还会失去输出的精确度。|74c3l|如果您确实需要long数的整数平方根，或者通常需要太大而无法准确表示为double的数字，请查看integer+square+root算法。|3979m|您还可以使用Guava库中的LongMath.sqrt()。|719is^0|0|9|A|6|Q|4|15|5|1D|6|2F|9|0|0|0|7|4|Z|6|1C|J|1|0|E|F|E|F|2|0^^$0|$1|$2|3|4|5|6|$7|8]]|9|$2|3|4|5|6|$7|A]]|B|$2|3|4|5|6|$7|C]]]|D|@$E|F|G|H|2|I|J|X|K|@$L|Y|M|Z|N|O]|$L|10|M|11|N|O]|$L|12|M|13|N|O]|$L|14|M|15|N|O]|$L|16|M|17|N|O]]|P|@$L|18|M|19|E|1A]]|6|$]]|$E|Q|G|R|2|I|J|1B|K|@]|P|@]|6|$]]|$E|S|G|T|2|I|J|1C|K|@$L|1D|M|1E|N|O]|$L|1F|M|1G|N|O]]|P|@$L|1H|M|1I|E|1J]]|6|$]]|$E|U|G|V|2|I|J|1K|K|@$L|1L|M|1M|N|O]]|P|@$L|1N|M|1O|E|1P]]|6|$]]|$E|W|G|-4|2|I|J|1Q|K|@]|P|@]|6|$]]]]

<code>Math.sqrt</code> operates on <code>double</code>s, not <code>long</code>s, so <code>mynum</code> gets converted to a <code>double</code> first. This is a 64-bits floating point number, which has "15–17 decimal digits of precision" (<a href="https://en.wikipedia.org/wiki/Double-precision_floating-point_format" rel="nofollow noreferrer">Wikipedia</a>).

Your input number has 16 digits, so you may be losing precision on the input already. You may also be losing precision on the output.

If you really need an integer square root of <code>long</code> numbers, or generally numbers that are too big for accurate representation as a <code>double</code>, look into <a href="https://en.wikipedia.org/wiki/Integer_square_root" rel="nofollow noreferrer">integer square root</a> algorithms.

You can also use <a href="https://github.com/google/guava/blob/master/guava/src/com/google/common/math/LongMath.java#L312" rel="nofollow noreferrer"><code>LongMath.sqrt()</code></a> from the Guava library.

entityMap|blocks|key|8en4c|text|double+ans+=+(double)Math.sqrt(15);
System.out.println("Double+:+"+%2B+ans);
double+ans_ans+=+ans+*+ans;
System.out.println("Double+:+"+%2B+ans_ans);

long+ans1+=+(long)Math.sqrt(15);
System.out.println("Long+:+"+%2B+ans1);
long+ans_ans1+=+ans1+*+ans1;
System.out.println("Long+:+"+%2B+ans_ans1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1h8mt|结果：|unstyled|cfi6i|Double+:+3.872983346207417
Double+:+15.000000000000002
Long+:+3
Long+:+9|3ls1b|我希望这能让你明白这一点。|3o6hv^0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$C|D]]|$2|E|4|F|6|G|8|N|9|@]|A|@]|B|$]]|$2|H|4|I|6|7|8|O|9|@]|A|@]|B|$C|D]]|$2|J|4|K|6|G|8|P|9|@]|A|@]|B|$]]|$2|L|4|-4|6|G|8|Q|9|@]|A|@]|B|$]]]]

<pre><code>double ans = (double)Math.sqrt(15);
System.out.println("Double : " + ans);
double ans_ans = ans * ans;
System.out.println("Double : " + ans_ans);

long ans1 = (long)Math.sqrt(15);
System.out.println("Long : " + ans1);
long ans_ans1 = ans1 * ans1;
System.out.println("Long : " + ans_ans1);
</code></pre>

Results : 

<pre><code>Double : 3.872983346207417
Double : 15.000000000000002
Long : 3
Long : 9
</code></pre>

I hope this makes it clear.

entityMap|blocks|key|358v8|text|答案是:四舍五入。(Long)Math.sqrt(7660142319573120)的结果是87522239，但数学结果是87522238,999999994287166259537761...。如果将ans值相乘，将其四舍五入以存储为整数，则得到的数字将大于确切结果的乘法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|8clv9^0|9|X|1A|8|1P|10|2T|3|0^^$0|$]|1|@$2|3|4|5|6|7|8|H|9|@$A|I|B|J|C|D]|$A|K|B|L|C|D]|$A|M|B|N|C|D]|$A|O|B|P|C|D]]|E|@]|F|$]]|$2|G|4|-4|6|7|8|Q|9|@]|E|@]|F|$]]]]

The answer is: rounding.
The result of <code>(Long)Math.sqrt(7660142319573120)</code> is <code>87522239</code>, but the mathematical result is <code>87522238,999999994287166259537761...</code>. 
if you multiply the <code>ans</code> value, which is rounded up in order to be stored as a whole number, you will get bigger number than multiplying the exact result.

entityMap|blocks|key|85k40|text|您不需要long类型，所有数字都可以在double中表示，Math.sqrt首先转换为double，然后通过FPU指令计算平方根(在标准PC上)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|elhgt|当a是范围内的整数时，数字b=a%5E2-1会出现这种情况|f3hvo|67108865+<=+a+<=+94906265|code-block|syntax|javascript|6rs86|b的平方根有一个级数展开，从|2jroh|a-1/(2*a)-1/(8*a%5E2)%2B...+|89dre|如果相对误差1/(2*a%5E2)低于机器ε，则最接近的可表示double号为a。|5el5q|另一方面，要使这个技巧起作用，需要a*a-1.0在double中完全可表示，这给出了条件|7bfa0|1/(2*a%5E2)+<mu=2%5E(-53)+<+1/(a%5E2)|eqqp3|或者|5qodl|2%5E52+<+a%5E2+<+2%5E53
2%5E26%2B1=67108865+<=+a+<=+floor(sqrt(2)*2%5E26)=94906265|9055p^0|J|6|T|9|0|1|1|D|7|0|0|0|1|0|0|6|9|T|6|11|1|0|H|7|P|6|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|12|9|@$A|13|B|14|C|D]|$A|15|B|16|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|17|9|@$A|18|B|19|C|D]|$A|1A|B|1B|C|D]]|E|@]|F|$]]|$2|I|4|J|6|K|8|1C|9|@]|E|@]|F|$L|M]]|$2|N|4|O|6|7|8|1D|9|@$A|1E|B|1F|C|D]]|E|@]|F|$]]|$2|P|4|Q|6|K|8|1G|9|@]|E|@]|F|$L|M]]|$2|R|4|S|6|7|8|1H|9|@$A|1I|B|1J|C|D]|$A|1K|B|1L|C|D]|$A|1M|B|1N|C|D]]|E|@]|F|$]]|$2|T|4|U|6|7|8|1O|9|@$A|1P|B|1Q|C|D]|$A|1R|B|1S|C|D]]|E|@]|F|$]]|$2|V|4|W|6|K|8|1T|9|@]|E|@]|F|$L|M]]|$2|X|4|Y|6|7|8|1U|9|@]|E|@]|F|$]]|$2|Z|4|10|6|K|8|1V|9|@]|E|@]|F|$L|M]]|$2|11|4|-4|6|7|8|1W|9|@]|E|@]|F|$]]]]

You do not need the long type, all numbers are representable in <code>double</code>, and <code>Math.sqrt</code> first converts to double then computes the square root via FPU instruction (on a standard PC). 

This situation occurs for numbers <code>b=a^2-1</code> where <code>a</code> is an integer in the range 

<pre><code>67108865 &lt;= a &lt;= 94906265
</code></pre>

The square root of <code>b</code> has a series expansion starting with

<pre><code>a-1/(2*a)-1/(8*a^2)+... 
</code></pre>

If the relative error <code>1/(2*a^2)</code> falls below the machine epsilon, the closest representable <code>double</code> number is <code>a</code>.

On the other hand for this trick to work one needs that <code>a*a-1.0</code> is exactly representable in <code>double</code>, which gives the conditions

<pre><code>1/(2*a^2) &lt;mu=2^(-53) &lt; 1/(a^2)
</code></pre>

or

<pre><code>2^52 &lt; a^2 &lt; 2^53
2^26+1=67108865 &lt;= a &lt;= floor(sqrt(2)*2^26)=94906265
</code></pre>

<pre><code>long mynum = Long.parseLong("7660142319573120");
long ans = (long)Math.sqrt(mynum) // output = 87522239
long ans_ans = ans * ans;
</code></pre>

In this case, I am getting ans_ans > mynum where it should be &lt;=mynum. Why such behavior? I tried this with node js as well. There also result is same.

Why Math.sqrt() outputs wrong value in java?

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋 

腾讯云代码助手

CODING DevOps

Cloud Studio

SDK中心

API中心

命令行工具

long mynum = Long.parseLong("7660142319573120");long ans = (long)Math.sqrt(mynum) // output = 87522239long ans_ans = ans * ans; 在本例中，我将在应该为<=mynum的位置获取ans_ans > mynum。为什么会有这样的行为？我也尝试过使用node js。结果也是一样

问为什么Math.sqrt()在java中输出错误的值？
EN

回答 5

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问为什么Math.sqrt()在java中输出错误的值？EN

回答 5

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问为什么Math.sqrt()在java中输出错误的值？
EN