blocks|key|1005163|text|C99有log2+(以及用于浮点型和long+double的log2f和log2l+)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1005164|entityMap|0|LINK|mutability|MUTABLE|url|http://www.opengroup.org/onlinepubs/009695399/functions/log2.html^0|4|4|U|5|10|5|4|4|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

C99 has <a href="http://www.opengroup.org/onlinepubs/009695399/functions/log2.html" rel="noreferrer"><code>log2</code></a> (as well as <code>log2f</code> and <code>log2l</code> for float and long double).

blocks|key|1219490|text|如果您正在寻找一个整数结果，您可以只确定值中设置的最高位，并返回其位置。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1219491|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

If you're looking for an integral result, you can just determine the highest bit set in the value and return its position.

blocks|key|1005230|text|#define+M_LOG2E+1.44269504088896340736+//+log2(e)

inline+long+double+log2(const+long+double+x){
++++return+log(x)+*+M_LOG2E;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1005231|(乘法可能比除法快)|unstyled|1005232|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>#define M_LOG2E 1.44269504088896340736 // log2(e)

inline long double log2(const long double x){
 return log(x) * M_LOG2E;
}
</code></pre>

(multiplication may be faster than division)

blocks|key|1219697|text|log2(int+n)+=+31+-+__builtin_clz(n)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1219698|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>log2(int n) = 31 - __builtin_clz(n)
</code></pre>

blocks|key|1219452|text|如http://en.wikipedia.org/wiki/Logarithm上所述|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1219453|logb(x)+=+logk(x)+/+logk(b)|code-block|syntax|javascript|1219454|这意味着：|1219455|log2(x)+=+log10(x)+/+log10(2)|1219456|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Logarithm^0|1|12|0|0|0|0|0^^$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]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

As stated on <a href="http://en.wikipedia.org/wiki/Logarithm" rel="noreferrer">http://en.wikipedia.org/wiki/Logarithm</a>:

<pre><code>logb(x) = logk(x) / logk(b)
</code></pre>

Which means that:

<pre><code>log2(x) = log10(x) / log10(2)
</code></pre>

blocks|key|1219443|text|log2(x)+=+log10(x)+/+log10(2)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1219444|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

blocks|key|1219564|text|uint16_t+log2(uint32_t+n)+{//but+truncated
+++++if+(n==0)+throw+...
+++++uint16_t+logValue+=+-1;
+++++while+(n)+{//
+++++++++logValue%2B%2B;
+++++++++n+>>=+1;
+++++}
+++++return+logValue;
+}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1219565|基本上和tomlogic的一样。|unstyled|offset|length|1219566|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/users/266392/tomlogic^0|0|4|8|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>uint16_t log2(uint32_t n) {//but truncated
 if (n==0) throw ...
 uint16_t logValue = -1;
 while (n) {//
 logValue++;
 n &gt;&gt;= 1;
 }
 return logValue;
 }
</code></pre>

Basically the same as <a href="https://stackoverflow.com/users/266392/tomlogic">tomlogic</a>'s.

blocks|key|1219683|text|你必须包括math.h+(C)或cmath+(C%2B%2B)，当然要记住，你必须遵循我们know...only+numbers>0的数学。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1219684|示例：|1219685|#include+<iostream>
#include+<cmath>
using+namespace+std;

int+main(){
++++cout<<log2(number);
}|code-block|syntax|javascript|1219686|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You have to include math.h (C) or cmath (C++)
Of course keep on mind that you have to follow the maths that we know...only numbers>0.

Example: 

<pre><code>#include &lt;iostream&gt;
#include &lt;cmath&gt;
using namespace std;

int main(){
 cout&lt;&lt;log2(number);
}
</code></pre>

blocks|key|1005384|text|我需要更高的精度，而不仅仅是最重要的位的位置，而且我使用的微控制器没有数学库。我发现，对于正整数值参数，只需使用2%5En值之间的线性近似就可以很好地工作。代码如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1005385|uint16_t+approx_log_base_2_N_times_256(uint16_t+n)
{
++++uint16_t+msb_only+=+0x8000;
++++uint16_t+exp+=+15;

++++if+(n+==+0)
++++++++return+(-1);
++++while+((n+&+msb_only)+==+0)+{
++++++++msb_only+>>=+1;
++++++++exp--;
++++}

++++return+(((uint16_t)((((uint32_t)+(n+%5E+msb_only))+<<+8)+/+msb_only))+%7C+(exp+<<+8));
}|code-block|syntax|javascript|1005386|在我的主程序中，我需要用一个整数结果来计算N*+log2(N)+/2：|1005387|temp+=((Uint32_t)+N)+*+approx_log_base_2_N_times_256)+/+512；|1005388|所有16位数值的偏差从未超过2%25|1005389|entityMap^0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I needed to have more precision that just the position of the most significant bit, and the microcontroller I was using had no math library. I found that just using a linear approximation between 2^n values for positive integer value arguments worked well. Here is the code:

<pre><code>uint16_t approx_log_base_2_N_times_256(uint16_t n)
{
 uint16_t msb_only = 0x8000;
 uint16_t exp = 15;

 if (n == 0)
 return (-1);
 while ((n &amp; msb_only) == 0) {
 msb_only &gt;&gt;= 1;
 exp--;
 }

 return (((uint16_t)((((uint32_t) (n ^ msb_only)) &lt;&lt; 8) / msb_only)) | (exp &lt;&lt; 8));
}
</code></pre>

In my main program, I needed to calculate N * log2(N) / 2 with an integer result:

temp = (((uint32_t) N) * approx_log_base_2_N_times_256) / 512;

and all 16 bit values were never off by more than 2%

blocks|key|1219861|text|以上所有答案都是正确的。如果有人需要它，我下面的答案可能会很有帮助。我在许多问题中都看到了这一要求，我们正在使用C语言解决这些问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1219862|log2+(x)+=+logy+(x)+/+logy+(2)|offset|length|style|CODE|1219863|但是，如果您使用的是C语言，并且希望得到整数形式的结果，则可以使用以下方法：|1219864|int+result+=+(int)(ceil(log(x)+/+log(2)));|code-block|syntax|javascript|1219865|希望这能有所帮助。|1219866|entityMap^0|0|0|U|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|X|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

All the above answers are correct. This answer of mine below can be helpful if someone needs it. I have seen this requirement in many questions which we are solving using C.
<code>log2 (x) = logy (x) / logy (2)</code>
However, if you are using C language and you want the result in integer, you can use the following:
<pre><code>int result = (int)(ceil(log(x) / log(2)));
</code></pre>
Hope this helps.

blocks|key|1005063|text|请参考您的基础数学课程log+n+/+log+2。在这种情况下，选择log还是log10都无关紧要，除以新基数的log就行了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1005064|entityMap^0|B|D|Y|3|13|5|1K|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

Consult your basic mathematics course, <code>log n / log 2</code>. It doesn't matter whether you choose <code>log</code> or <code>log10</code>in this case, dividing by the <code>log</code> of the new base does the trick.

blocks|key|1005459|text|Ustaman的改进版本|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1005460|static+inline+uint64_t
log2(uint64_t+n)
{
++++uint64_t+val;
++++for+(val+=+0;+n+>+1;+val%2B%2B,+n+>>=+1);

++++return+val;
}|code-block|syntax|javascript|1005461|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Improved version of what Ustaman Sangat did

<pre><code>static inline uint64_t
log2(uint64_t n)
{
 uint64_t val;
 for (val = 0; n &gt; 1; val++, n &gt;&gt;= 1);

 return val;
}
</code></pre>

Is there any way to write log(base 2) function?

The C language has 2 built in function -->>

1.<code>log</code> which is base e.

2.<code>log10</code> base 10;

But I need log function of base 2.How to calculate this.

How to write log base(2) in c/c++

有没有办法写log(基数为2)函数？C语言有2个内置函数-->>1.以e为基础的log。2.log10碱基10；但是我需要以2为底的对数函数，如何计算。

问如何用c/c++编写log base(2)
EN

回答 12

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