blocks|key|1465201|text|你只需要创建一个面具：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465202|int+createMask(int+a,+int+b){
int+c+=+a;
int+mask+=+0;
/*First+we+set+the+lenght+of+the+mask*/
while(c+<=+b){++/*Including+b*/
mask+<<=+1;
mask+=+mask%7C1;
c%2B%2B;
}
/*Then+we+set+the+position+to+the+mask,+the+first+bit+is+in+the+position+0*/
c=0;
while(c<a){
c%2B%2B;
mask+<<=+1+;

}
return+mask;
}|code-block|syntax|javascript|1465203|我还没有测试这个功能，但它只是为了解释制作面具的方法。|1465204|最后的功能可能是这样的：|1465205|int+getBits(int+value,+int+a,+int+b){
int+mask+=+createMask(a,+b);
mask+&=+value;
//Now+we+have+to+move+the+bits+to+the+right
while(a>0){
mask+>>=+1;
a--;
}
return+mask;

}|1465206|举个例子，如果你想要前6位，你必须编码:+getBits(myValue，0，5)。|1465207|我不知道你对签名人和未签名人的数字是什么意思，但我希望它能对你有所帮助。|1465208|学我的英语。|1465209|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

You just have to create a mask:

<pre><code>int createMask(int a, int b){
int c = a;
int mask = 0;
/*First we set the lenght of the mask*/
while(c &lt;= b){ /*Including b*/
mask &lt;&lt;= 1;
mask = mask|1;
c++;
}
/*Then we set the position to the mask, the first bit is in the position 0*/
c=0;
while(c&lt;a){
c++;
mask &lt;&lt;= 1 ;

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

I haven't tested the function yet, but its just for explaining a way to make a mask.

And the final function may be something like this:

<pre><code>int getBits(int value, int a, int b){
int mask = createMask(a, b);
mask &amp;= value;
//Now we have to move the bits to the right
while(a&gt;0){
mask &gt;&gt;= 1;
a--;
}
return mask;

}
</code></pre>

An example, if you want the first 6 bits, you have to code: getBits(myValue, 0, 5).

Im not sure what did you mean about the signeds and unsigneds numbers, but i hope it can help you.

Srry for my english.

blocks|key|1465269|text|我怀疑你可能想用另一种方式来处理整个问题。与其提取比特，为什么不直接使用比特掩码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465270|例如，要检查是否启用了字节中最重要的位：|1465271|if(byte+&+0xf0)+{}|code-block|syntax|javascript|1465272|要检查最不重要的位数，应该是：|1465273|if(byte+&+0x01)+{}|1465274|要检查多个比特(或一个“范围”)，例如低阶咬取：|1465275|if(byte+&+0x0f)+{}|1465276|从你所说的来看，我怀疑这更接近你想要的东西，而且比转移到提取比特要简单得多。|1465277|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

I suspect you might want to approach the entire problem in a different way. Rather than extracting bits, why not just use bit masks.

For example, to check if the most significant bit in a byte is enabled:

<pre><code>if(byte &amp; 0xf0) {}
</code></pre>

To check for the least significant bit it would be:

<pre><code>if(byte &amp; 0x01) {}
</code></pre>

To check for multiple (or a "range") of bits, say the low order nibble:

<pre><code>if(byte &amp; 0x0f) {}
</code></pre>

From what you've said, I suspect this is much closer to what you want and much simpler than shifting to extract bits.

blocks|key|970045|text|这有点有趣:)分三个简单的步骤：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|970046|将您的值按lo值右移，并使用lo降低hi。这将问题简化为“获取最低的hi位”。|ordered-list-item|offset|length|style|CODE|970047|剪辑最高位-一个定制的掩码是创建的动态。|970048|如果有必要，使用最高位来签名-扩展结果(基于Sign+extending+from+a+constant+bit+width+in+C#的篡改)。|970049|970050|970051|我不知道建议函数原型的原因，但我建议使用order+lo,+hi而不是hi,+lo。不知怎么的，10,14比相反的感觉更自然，尽管当计算从左到右时，位数从高到低--计算机应该使我们的事情变得更容易！|970052|#include+<stdio.h>
#include+<stdbool.h>

int+getField+(int+value,+int+hi,+int+lo,+bool+isSigned)
{
++++/*+step+1:+clip+off+lower+bits+*/
++++value+>>=+lo;
++++hi+-=+lo-1;
++++/*+step+2:+clip+off+higher+bits+*/
++++value+&=+~(-1<<hi);
++++/*+step+3:+extend+sign+*/
++++if+(isSigned+&&+(value+&+(1<<(hi-1))))
++++++++value+%7C=+-(1<<hi);
++++return+value;
}

int+main+(void)
{
++++int+i;
++++i+=+getField+(0x123456c8,+14,10,+true);
++++printf+("i+=+%25d+/+%25Xh\n",+i,i);
++++return+0;
}|code-block|syntax|javascript|970053|结果：|970054|i+=+-11+/+FFFFFFF5h|970055|哪个是正确的位集：|970056|++16+++12++++8++++4++++0+<-+bit+position
...4++++5++++6++++7++++8+<-+value
0100+0101+0110+0111+1000+<-+bitwise
++++++---+--+++++++++++++<-+mask
++++++101+01+++++++++++++<-+result
+..111101+01+++++++++++++<+sign+extended+result|970057|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/3981106/sign-extending-from-a-constant-bit-width-in-c-sharp^0|0|5|2|E|2|I|2|0|0|M|1A|0|0|0|0|Q|6|Z|6|1C|5|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1C|8|@$E|1D|F|1E|G|H]|$E|1F|F|1G|G|H]|$E|1H|F|1I|G|H]]|9|@]|A|$]]|$1|I|3|J|5|D|7|1J|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|1K|8|@]|9|@$E|1L|F|1M|1|1N]]|A|$]]|$1|M|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1Q|8|@$E|1R|F|1S|G|H]|$E|1T|F|1U|G|H]|$E|1V|F|1W|G|H]]|9|@]|A|$]]|$1|Q|3|R|5|S|7|1X|8|@]|9|@]|A|$T|U]]|$1|V|3|W|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|X|3|Y|5|S|7|1Z|8|@]|9|@]|A|$T|U]]|$1|Z|3|10|5|6|7|20|8|@]|9|@]|A|$]]|$1|11|3|12|5|S|7|21|8|@]|9|@]|A|$T|U]]|$1|13|3|-4|5|6|7|22|8|@]|9|@]|A|$]]]|14|$15|$5|16|17|18|A|$19|1A]]]]

That was a bit of fun :) In three easy steps:

<ol>
<li>shift your value right by the amount <code>lo</code> and decrease <code>hi</code> with <code>lo</code>. This simplifies the problem to 'get the lowest hi bits'.</li>
<li>clip off the highest bits -- a custom mask is created on the fly.</li>
<li>if necessary, use the highest bit to sign-extend the result (bitfiddling based on <a href="https://stackoverflow.com/questions/3981106/sign-extending-from-a-constant-bit-width-in-c-sharp">Sign extending from a constant bit width in C#</a>).</li>
</ol>

I don't know the reason for the suggested function prototype, but I would suggest using the order <code>lo, hi</code> rather than <code>hi, lo</code>. Somehow <code>10,14</code> feels more natural than the other way around, even though bits count down from high to low, when counted left to right -- the computer is supposed to make things easier for us!

<pre><code>#include &lt;stdio.h&gt;
#include &lt;stdbool.h&gt;

int getField (int value, int hi, int lo, bool isSigned)
{
 /* step 1: clip off lower bits */
 value &gt;&gt;= lo;
 hi -= lo-1;
 /* step 2: clip off higher bits */
 value &amp;= ~(-1&lt;&lt;hi);
 /* step 3: extend sign */
 if (isSigned &amp;&amp; (value &amp; (1&lt;&lt;(hi-1))))
 value |= -(1&lt;&lt;hi);
 return value;
}

int main (void)
{
 int i;
 i = getField (0x123456c8, 14,10, true);
 printf ("i = %d / %Xh\n", i,i);
 return 0;
}
</code></pre>

Result:

<pre><code>i = -11 / FFFFFFF5h
</code></pre>

which is the correct bit set:

<pre class="lang-none prettyprint-override"><code> 16 12 8 4 0 &lt;- bit position
...4 5 6 7 8 &lt;- value
0100 0101 0110 0111 1000 &lt;- bitwise
 --- -- &lt;- mask
 101 01 &lt;- result
 ..111101 01 &lt; sign extended result
</code></pre>

I have a function to extract a single bit from a number:

<pre><code>int getBit (int value, int position) {

 return value &amp; (1 &lt;&lt; position));

 }
</code></pre>

But how do I do it for a range (both for signed and unsigned numbers)? For instance:

get bits 10:14 from 0x12345678 (signed 0) = 0x15

<pre><code>int getField (int value, int hi, int lo, bool isSigned)
</code></pre>

How to return multiple bits from a number in C

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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我有一个函数从一个数字中提取一个位：int getBit (int value, int position) { return value & (1 << position)); }但是如何对一个范围(包括有符号数字和无符号数字)这样做呢？例如：从0x12345678 (签名0) = 0x15获得10:14...

问如何从C中的数字返回多个位
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问如何从C中的数字返回多个位EN