blocks|key|4374192|text|检查是否设置了位N(从0开始)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4374193|temp+&+(1+<<+N)|code-block|syntax|javascript|4374194|这方面没有内置的函数。|4374195|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Check if bit N (starting from 0) is set:

<pre><code>temp &amp; (1 &lt;&lt; N)
</code></pre>

There is no builtin function for this.

blocks|key|1022484|text|如果是C%2B%2B，我只会使用std::bitset。很简单。直截了当。没有机会犯愚蠢的错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022485|typedef+std::bitset<sizeof(int)>+IntBits;
bool+is_set+=+IntBits(value).test(position);|code-block|syntax|javascript|1022486|或者说这种愚蠢的行为|1022487|template<unsigned+int+Exp>
struct+pow_2+{
++++static+const+unsigned+int+value+=+2+*+pow_2<Exp-1>::value;
};

template<>
struct+pow_2<0>+{
++++static+const+unsigned+int+value+=+1;
};

template<unsigned+int+Pos>
bool+is_bit_set(unsigned+int+value)
{
++++return+(value+&+pow_2<Pos>::value)+!=+0;
}+

bool+result+=+is_bit_set<2>(value);|1022488|entityMap^0|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I would just use a std::bitset if it's C++. Simple. Straight-forward. No chance for stupid errors.

<pre><code>typedef std::bitset&lt;sizeof(int)&gt; IntBits;
bool is_set = IntBits(value).test(position);
</code></pre>

or how about this silliness

<pre><code>template&lt;unsigned int Exp&gt;
struct pow_2 {
 static const unsigned int value = 2 * pow_2&lt;Exp-1&gt;::value;
};

template&lt;&gt;
struct pow_2&lt;0&gt; {
 static const unsigned int value = 1;
};

template&lt;unsigned int Pos&gt;
bool is_bit_set(unsigned int value)
{
 return (value &amp; pow_2&lt;Pos&gt;::value) != 0;
} 

bool result = is_bit_set&lt;2&gt;(value);
</code></pre>

blocks|key|4374740|text|所选择的答案实际上是错误的。下面的函数将返回位的位置或0，具体取决于该位是否实际启用。这不是发帖者所要求的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4374741|#define+CHECK_BIT(var,pos)+((var)+&+(1<<(pos)))|code-block|syntax|javascript|4374742|这就是发帖者最初想要的。如果使能位而不是位置，则下面的函数将返回1或0。|4374743|#define+CHECK_BIT(var,pos)+(((var)>>(pos))+&+1)|4374744|entityMap^0|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

What the selected answer is doing is actually wrong. The below function will return the bit position or 0 depending on if the bit is actually enabled. This is not what the poster was asking for.

<pre><code>#define CHECK_BIT(var,pos) ((var) &amp; (1&lt;&lt;(pos)))
</code></pre>

Here is what the poster was originally looking for. The below function will return either a 1 or 0 if the bit is enabled and not the position.

<pre><code>#define CHECK_BIT(var,pos) (((var)&gt;&gt;(pos)) &amp; 1)
</code></pre>

blocks|key|3526972|text|是啊，我知道我不能“”这样做。但我通常会这样写：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3526973|++++/*+Return+type+(8/16/32/64+int+size)+is+specified+by+argument+size.+*/
template<class+TYPE>+inline+TYPE+BIT(const+TYPE+&+x)
{+return+TYPE(1)+<<+x;+}

template<class+TYPE>+inline+bool+IsBitSet(const+TYPE+&+x,+const+TYPE+&+y)
{+return+0+!=+(x+&+y);+}|code-block|syntax|javascript|3526974|例如：|3526975|IsBitSet(+foo,+BIT(3)+%7C+BIT(6)+);++//+Checks+if+Bit+3+OR+6+is+set.|3526976|在其他方面，这种方法：|3526977|在没有my+knowledge和integers.|3526978|Detects模板的情况下，consent.|unordered-list-item|3526979|
|3526980|Inlined+|3526981|3526982|可容纳8/16/32/64位overhead.|3526983|3526984|const&+IsBitSet(int32，int64)调用，因此没有调用overhead.|3526985|3526986|const&引用的函数，因此没有任何需要复制/复制。我们可以保证，编译器会发现任何试图更改arguments.|3526987|3526988|0!=的拼写错误，从而使代码更清晰、更明显。编写代码的主要要点始终是与其他程序员进行清晰而有效的沟通，包括那些技能较低的程序员。|3526989|3526990|虽然不适用于这种特定的情况...一般而言，模板化函数避免了多次计算参数的问题。一些#define宏的已知问题。例如：#define+ABS(X)+(X)+<0)？-(X)：(X))|3526991|3526992|+ABS(i%2B%2B);|3526993|entityMap^0|A|1|0|0|0|0|0|0|0|0|0|0|0|0|0|6|0|0|0|6|I|7|0|0|0|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1J|8|@$9|1K|A|1L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1M|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1N|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1O|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1P|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1Q|8|@]|D|@]|E|$]]|$1|S|3|T|5|U|7|1R|8|@]|D|@]|E|$]]|$1|V|3|W|5|6|7|1S|8|@]|D|@]|E|$]]|$1|X|3|Y|5|U|7|1T|8|@]|D|@]|E|$]]|$1|Z|3|W|5|6|7|1U|8|@]|D|@]|E|$]]|$1|10|3|11|5|U|7|1V|8|@]|D|@]|E|$]]|$1|12|3|W|5|6|7|1W|8|@]|D|@]|E|$]]|$1|13|3|14|5|U|7|1X|8|@$9|1Y|A|1Z|B|C]]|D|@]|E|$]]|$1|15|3|W|5|6|7|20|8|@]|D|@]|E|$]]|$1|16|3|17|5|U|7|21|8|@$9|22|A|23|B|C]|$9|24|A|25|B|C]]|D|@]|E|$]]|$1|18|3|W|5|6|7|26|8|@]|D|@]|E|$]]|$1|19|3|1A|5|U|7|27|8|@$9|28|A|29|B|C]]|D|@]|E|$]]|$1|1B|3|W|5|6|7|2A|8|@]|D|@]|E|$]]|$1|1C|3|1D|5|U|7|2B|8|@]|D|@]|E|$]]|$1|1E|3|-4|5|6|7|2C|8|@]|D|@]|E|$]]|$1|1F|3|1G|5|6|7|2D|8|@]|D|@]|E|$]]|$1|1H|3|-4|5|6|7|2E|8|@]|D|@]|E|$]]]|1I|$]]

Yeah, I know I don't "have" to do it this way. But I usually write:

<pre><code> /* Return type (8/16/32/64 int size) is specified by argument size. */
template&lt;class TYPE&gt; inline TYPE BIT(const TYPE &amp; x)
{ return TYPE(1) &lt;&lt; x; }

template&lt;class TYPE&gt; inline bool IsBitSet(const TYPE &amp; x, const TYPE &amp; y)
{ return 0 != (x &amp; y); }
</code></pre>

E.g.:

<pre><code>IsBitSet( foo, BIT(3) | BIT(6) ); // Checks if Bit 3 OR 6 is set.
</code></pre>

Amongst other things, this approach:

<ul>
<li>Accommodates 8/16/32/64 bit integers.</li>
<li>Detects IsBitSet(int32,int64) calls without my knowledge &amp; consent.</li>
<li>Inlined Template, so no function calling overhead.</li>
<li>const&amp; references, so nothing needs to be duplicated/copied. And we are guaranteed that the compiler will pick up any typo's that attempt to change the arguments.</li>
<li>0!= makes the code more clear &amp; obvious. The primary point to writing code is always to communicate clearly and efficiently with other programmers, including those of lesser skill.</li>
<li>While not applicable to this particular case... In general, templated functions avoid the issue of evaluating arguments multiple times. A known problem with some #define macros. E.g.: #define ABS(X) (((X)&lt;0) ? - (X) : (X)) &nbsp; &nbsp; &nbsp; ABS(i++);</li>
</ul>

blocks|key|3526896|text|根据this+description+of+bit-fields的说法，有一种直接定义和访问字段的方法。此条目中的示例如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3526897|struct+preferences+{
++++unsigned+int+likes_ice_cream+:+1;
++++unsigned+int+plays_golf+:+1;
++++unsigned+int+watches_tv+:+1;
++++unsigned+int+reads_books+:+1;
};+

struct+preferences+fred;

fred.likes_ice_cream+=+1;
fred.plays_golf+=+1;
fred.watches_tv+=+1;
fred.reads_books+=+0;

if+(fred.likes_ice_cream+==+1)
++++/*+...+*/|code-block|syntax|javascript|3526898|此外，还有一个警告：|3526899|3526900|然而，结构中的位成员有实际的缺点。首先，内存中位的顺序依赖于体系结构，并且不同编译器的内存填充规则也不同。此外，许多流行的编译器生成低效的代码来读取和写入位成员，并且存在与位字段相关的潜在的严重线程安全问题(特别是在多处理器系统上)，这是因为大多数机器不能在内存中操作任意位集，而是必须加载和存储整个字。|blockquote|3526901|3526902|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Bit_field^0|2|U|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|E|5|F|7|11|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|12|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|13|8|@]|9|@]|C|$]]|$1|L|3|M|5|N|7|14|8|@]|9|@]|C|$]]|$1|O|3|-4|5|6|7|15|8|@]|9|@]|C|$]]|$1|P|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|Q|$R|$5|S|T|U|C|$V|W]]]]

According to <a href="http://en.wikipedia.org/wiki/Bit_field" rel="noreferrer">this description of bit-fields</a>, there is a method for defining and accessing fields directly. The example in this entry goes:

<pre><code>struct preferences {
 unsigned int likes_ice_cream : 1;
 unsigned int plays_golf : 1;
 unsigned int watches_tv : 1;
 unsigned int reads_books : 1;
}; 

struct preferences fred;

fred.likes_ice_cream = 1;
fred.plays_golf = 1;
fred.watches_tv = 1;
fred.reads_books = 0;

if (fred.likes_ice_cream == 1)
 /* ... */
</code></pre>

Also, there is a warning there:

<blockquote>
 However, bit members in structs have practical drawbacks. First, the ordering of bits in memory is architecture dependent and memory padding rules varies from compiler to compiler. In addition, many popular compilers generate inefficient code for reading and writing bit members, and there are potentially severe thread safety issues relating to bit fields (especially on multiprocessor systems) due to the fact that most machines cannot manipulate arbitrary sets of bits in memory, but must instead load and store whole words.
</blockquote>

blocks|key|1022331|text|可以使用Bitset+-+http://www.cppreference.com/wiki/stl/bitset/start。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1022332|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cppreference.com/wiki/stl/bitset/start^0|D|1D|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

You can use a Bitset - <a href="http://www.cppreference.com/wiki/stl/bitset/start" rel="noreferrer">http://www.cppreference.com/wiki/stl/bitset/start</a>.

blocks|key|1022499|text|使用std::bitset|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022500|#include+<bitset>
#include+<iostream>

int+main()
{
++++int+temp+=+0x5E;
++++std::bitset<sizeof(int)*CHAR_BITS>+++bits(temp);

++++//+0+->+bit+1
++++//+2+->+bit+3
++++std::cout+<<+bits[2]+<<+std::endl;
}|code-block|syntax|javascript|1022501|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|$]]

Use std::bitset

<pre><code>#include &lt;bitset&gt;
#include &lt;iostream&gt;

int main()
{
 int temp = 0x5E;
 std::bitset&lt;sizeof(int)*CHAR_BITS&gt; bits(temp);

 // 0 -&gt; bit 1
 // 2 -&gt; bit 3
 std::cout &lt;&lt; bits[2] &lt;&lt; std::endl;
}
</code></pre>

blocks|key|1017441|text|我试图读取一个32位整数，它为PDF中的对象定义了标志，但这对我不起作用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1017442|是什么修复了它改变了定义：|1017443|#define+CHECK_BIT(var,pos)+((var+&+(1+<<+pos))+==+(1+<<+pos))|code-block|syntax|javascript|1017444|操作数&返回一个带有标志的整数，这两个标志都在1中，但它没有正确地转换为布尔值，这就解决了问题|1017445|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

i was trying to read a 32-bit integer which defined the flags for an object in PDFs and this wasn't working for me

what fixed it was changing the define:

<pre><code>#define CHECK_BIT(var,pos) ((var &amp; (1 &lt;&lt; pos)) == (1 &lt;&lt; pos))
</code></pre>

the operand &amp; returns an integer with the flags that both have in 1, and it wasn't casting properly into boolean, this did the trick

blocks|key|1022611|text|我使用的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022612|#define+CHECK_BIT(var,pos)+(+(((var)+&+(pos))+>+0+)+?+(1)+:+(0)+)|code-block|syntax|javascript|1022613|其中"pos“定义为2%5En+(i.g.1,2,4,8,16,32+...)|1022614|返回:如果为true，则返回1；如果为false，则返回0|1022615|entityMap^0|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I use this:

<pre><code>#define CHECK_BIT(var,pos) ( (((var) &amp; (pos)) &gt; 0 ) ? (1) : (0) )
</code></pre>

where "pos" is defined as 2^n (i.g. 1,2,4,8,16,32 ...)

Returns: 
 1 if true
 0 if false

blocks|key|309205|text|有，即_bittest内部指令。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|309206|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/h65k4tze(VS.80).aspx^0|3|8|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

There is, namely the <a href="http://msdn.microsoft.com/en-us/library/h65k4tze(VS.80).aspx" rel="nofollow noreferrer">_bittest</a> intrinsic instruction.

blocks|key|1017398|text|对于低级特定于x86的解决方案，请使用x86+TEST操作码。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1017399|不过，您的编译器应该将_bittest转换为...|1017400|entityMap|0|LINK|mutability|MUTABLE|url|http://www.itis.mn.it/linux/quarta/x86/test.htm|1|http://msdn.microsoft.com/en-us/library/h65k4tze(VS.80).aspx^0|N|4|0|0|B|8|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|F|3|-4|5|6|7|X|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]|N|$5|I|J|K|C|$L|O]]]]

For the low-level x86 specific solution use the x86 <a href="http://www.itis.mn.it/linux/quarta/x86/test.htm" rel="nofollow noreferrer">TEST</a> opcode.

Your compiler should turn <a href="http://msdn.microsoft.com/en-us/library/h65k4tze(VS.80).aspx" rel="nofollow noreferrer">_bittest</a> into this though...

blocks|key|309178|text|你可以“模拟”移位和掩码:如果((0x5e/(2*2*2))%252)+...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|309179|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|$]]

You could "simulate" shifting and masking: if((0x5e/(2*2*2))%2) ...

blocks|key|4374698|text|为什么不使用像这样简单的东西呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4374699|uint8_t+status+=+255;
cout+<<+"binary:+";

for+(int+i=((sizeof(status)*8)-1);+i>-1;+i--)
{
++if+((status+&+(1+<<+i)))
++{
++++cout+<<+"1";
++}+
++else
++{
++++cout+<<+"0";
++}
}|code-block|syntax|javascript|4374700|输出:二进制：11111111|offset|length|style|CODE|4374701|entityMap^0|0|0|7|8|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Why not use something as simple as this?

<pre><code>uint8_t status = 255;
cout &lt;&lt; "binary: ";

for (int i=((sizeof(status)*8)-1); i&gt;-1; i--)
{
 if ((status &amp; (1 &lt;&lt; i)))
 {
 cout &lt;&lt; "1";
 } 
 else
 {
 cout &lt;&lt; "0";
 }
}
</code></pre>

OUTPUT: binary: <code>11111111</code>

blocks|key|4374758|text|一种方法是在以下条件下进行检查：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4374759|if+(+(mask+>>+bit+)+&+1)|code-block|syntax|javascript|4374760|解释程序将是：|4374761|#include+<stdio.h>

unsigned+int+bitCheck(unsigned+int+mask,+int+pin);

int+main(void){
+++unsigned+int+mask+=+6;++//+6+=+0110
+++int+pin0+=+0;
+++int+pin1+=+1;
+++int+pin2+=+2;
+++int+pin3+=+3;
+++unsigned+int+bit0=+bitCheck(+mask,+pin0);
+++unsigned+int+bit1=+bitCheck(+mask,+pin1);
+++unsigned+int+bit2=+bitCheck(+mask,+pin2);
+++unsigned+int+bit3=+bitCheck(+mask,+pin3);

+++printf("Mask+=+%25d+==>>++0110\n",+mask);

+++if+(+bit0+==+1+){
++++++printf("Pin+%25d+is+Set\n",+pin0);
+++}else{
++++++printf("Pin+%25d+is+not+Set\n",+pin0);
+++}

++++if+(+bit1+==+1+){
++++++printf("Pin+%25d+is+Set\n",+pin1);
+++}else{
++++++printf("Pin+%25d+is+not+Set\n",+pin1);
+++}

+++if+(+bit2+==+1+){
++++++printf("Pin+%25d+is+Set\n",+pin2);
+++}else{
++++++printf("Pin+%25d+is+not+Set\n",+pin2);
+++}

+++if+(+bit3+==+1+){
++++++printf("Pin+%25d+is+Set\n",+pin3);
+++}else{
++++++printf("Pin+%25d+is+not+Set\n",+pin3);
+++}
}

unsigned+int+bitCheck(unsigned+int+mask,+int+bit){
+++if+(+(mask+>>+bit+)+&+1){
++++++return+1;
+++}else{
++++++return+0;
+++}
}|4374762|输出：|4374763|4374764|掩码=6+==>>+0110引脚0未设置引脚1设置引脚2设置引脚3未设置|blockquote|4374765|4374766|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|V|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|X|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|Z|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|S|$]]

One approach will be checking within the following condition:

<pre><code>if ( (mask &gt;&gt; bit ) &amp; 1)
</code></pre>

An explanation program will be:

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

unsigned int bitCheck(unsigned int mask, int pin);

int main(void){
 unsigned int mask = 6; // 6 = 0110
 int pin0 = 0;
 int pin1 = 1;
 int pin2 = 2;
 int pin3 = 3;
 unsigned int bit0= bitCheck( mask, pin0);
 unsigned int bit1= bitCheck( mask, pin1);
 unsigned int bit2= bitCheck( mask, pin2);
 unsigned int bit3= bitCheck( mask, pin3);

 printf("Mask = %d ==&gt;&gt; 0110\n", mask);

 if ( bit0 == 1 ){
 printf("Pin %d is Set\n", pin0);
 }else{
 printf("Pin %d is not Set\n", pin0);
 }

 if ( bit1 == 1 ){
 printf("Pin %d is Set\n", pin1);
 }else{
 printf("Pin %d is not Set\n", pin1);
 }

 if ( bit2 == 1 ){
 printf("Pin %d is Set\n", pin2);
 }else{
 printf("Pin %d is not Set\n", pin2);
 }

 if ( bit3 == 1 ){
 printf("Pin %d is Set\n", pin3);
 }else{
 printf("Pin %d is not Set\n", pin3);
 }
}

unsigned int bitCheck(unsigned int mask, int bit){
 if ( (mask &gt;&gt; bit ) &amp; 1){
 return 1;
 }else{
 return 0;
 }
}
</code></pre>

Output:

<blockquote>
<pre><code>Mask = 6 ==&gt;&gt; 0110
Pin 0 is not Set
Pin 1 is Set
Pin 2 is Set
Pin 3 is not Set
</code></pre>
</blockquote>

blocks|key|3527449|text|#define+CHECK_BIT(var,pos)+((var>>pos)+&+1)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3527450|位置位位置从0开始。|unstyled|3527451|返回0或1。|3527452|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>#define CHECK_BIT(var,pos) ((var&gt;&gt;pos) &amp; 1)
</code></pre>

pos - Bit position strarting from 0.

returns 0 or 1.

blocks|key|309321|text|如果你只是想要一种真正的硬编码方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|309322|+#define+IS_BIT3_SET(var)+(+((var)+&+0x04)+==+0x04+)|code-block|syntax|javascript|309323|请注意，此hw依赖于hw，并假定此位顺序为7654+3210，var为8位。|309324|#include+"stdafx.h"
#define+IS_BIT3_SET(var)+(+((var)+&+0x04)+==+0x04+)
int+_tmain(int+argc,+_TCHAR*+argv[])
{
++++int+temp+=0x5E;
++++printf("+%25d+\n",+IS_BIT3_SET(temp));
++++temp+=+0x00;
++++printf("+%25d+\n",+IS_BIT3_SET(temp));
++++temp+=+0x04;
++++printf("+%25d+\n",+IS_BIT3_SET(temp));
++++temp+=+0xfb;
++++printf("+%25d+\n",+IS_BIT3_SET(temp));
++++scanf("waitng+%25d",&temp);

++++return+0;
}|309325|结果如下：|309326|1+0+1+0|309327|entityMap^0|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

if you just want a real hard coded way:

<pre><code> #define IS_BIT3_SET(var) ( ((var) &amp; 0x04) == 0x04 )
</code></pre>

note this hw dependent and assumes this bit order 7654 3210 and var is 8 bit.

<pre><code>#include "stdafx.h"
#define IS_BIT3_SET(var) ( ((var) &amp; 0x04) == 0x04 )
int _tmain(int argc, _TCHAR* argv[])
{
 int temp =0x5E;
 printf(" %d \n", IS_BIT3_SET(temp));
 temp = 0x00;
 printf(" %d \n", IS_BIT3_SET(temp));
 temp = 0x04;
 printf(" %d \n", IS_BIT3_SET(temp));
 temp = 0xfb;
 printf(" %d \n", IS_BIT3_SET(temp));
 scanf("waitng %d",&amp;temp);

 return 0;
}
</code></pre>

Results in:

1
0
1
0

blocks|key|309537|text|虽然现在回答已经很晚了，但有一种简单的方法可以发现是否设置了第N位，只需使用幂和模数学运算符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|309538|假设我们想知道'temp‘是否设置了第N位。如果设置了位，则下面的布尔表达式将返回true，否则返回0。|309539|309540|(+temp+MODULUS+2%5EN%2B1+>=+2%5EN+)|unordered-list-item|309541|309542|考虑以下示例：|309543|309544|int+temp+=+0x5E；//二进制0b1011110+//位0+is+LSB|309545|309546|如果我想知道是否设置了第三位，我会得到|309547|309548|(94模数16)+=+14+>+2%5E3|309549|309550|因此，表达式返回true，表示设置了第三位。|309551|entityMap^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Z|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|11|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|14|8|@]|9|@]|A|$]]|$1|L|3|M|5|G|7|15|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|16|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|17|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|R|3|S|5|G|7|19|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1B|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|X|$]]

While it is quite late to answer now, there is a simple way one could find if Nth bit is set or not, simply using POWER and MODULUS mathematical operators.

Let us say we want to know if 'temp' has Nth bit set or not. The following boolean expression will give true if bit is set, 0 otherwise.

<ul>
<li>( temp MODULUS 2^N+1 >= 2^N )</li>
</ul>

Consider the following example:

<ul>
<li>int temp = 0x5E; // in binary 0b1011110 // BIT 0 is LSB</li>
</ul>

If I want to know if 3rd bit is set or not, I get

<ul>
<li>(94 MODULUS 16) = 14 > 2^3</li>
</ul>

So expression returns true, indicating 3rd bit is set.

blocks|key|1022652|text|为什么所有这些位移位操作和库函数都需要？如果你有OP发布的值:+1011110，并且你想知道是否设置了右数第三个位置的位，只需执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1022653|int+temp+=+0b1011110;
if(+temp+&+4+)+++/*+or+(temp+&+0b0100)+if+that's+how+you+roll+*/
++DoSomething();|code-block|syntax|javascript|1022654|或者，更容易被代码的未来读者解释的东西，而不需要#include：|1022655|int+temp+=+0b1011110;
_Bool+bThirdBitIsSet+=+(temp+&+4)+?+1+:+0;
if(+bThirdBitIsSet+)
++DoSomething();|1022656|或者如果你想让它看起来更漂亮一点：|1022657|#include+<stdbool.h>
int+temp+=+0b1011110;
bool+bThirdBitIsSet+=+(temp+&+4)+?+true+:+false;
if(+bThirdBitIsSet+)
++DoSomething();|1022658|entityMap^0|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Why all these bit shifting operations and need for library functions? If you have the value the OP posted: 1011110 and you want to know if the bit in the 3rd position from the right is set, just do:
<pre><code>int temp = 0b1011110;
if( temp &amp; 4 ) /* or (temp &amp; 0b0100) if that's how you roll */
 DoSomething();
</code></pre>
Or, something that may be more easily interpreted by future readers of the code with no #include needed:
<pre><code>int temp = 0b1011110;
_Bool bThirdBitIsSet = (temp &amp; 4) ? 1 : 0;
if( bThirdBitIsSet )
 DoSomething();
</code></pre>
Or if you like it to look a bit prettier:
<pre><code>#include &lt;stdbool.h&gt;
int temp = 0b1011110;
bool bThirdBitIsSet = (temp &amp; 4) ? true : false;
if( bThirdBitIsSet )
 DoSomething();
</code></pre>

blocks|key|309785|text|先例答案向您展示了如何处理位校验，但更常见的情况是，它都是关于以整数编码的标志，这在任何先例案例中都没有很好的定义。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|309786|在一个典型的场景中，标志被定义为整数本身，对于它引用的特定位，位被定义为1。在下面的示例中，您可以检查整数是否有标志列表中的任何标志(多个错误标志连接在一起)，或者每个标志是否都在整数中(多个成功标志连接在一起)。|309787|下面是一个如何处理整数中的标志的示例。|309788|这里提供了一个现场示例：https://rextester.com/XIKE82408|offset|length|309789|//g%2B%2B++7.4.0

#include+<iostream>
#include+<stdint.h>

inline+bool+any_flag_present(unsigned+int+value,+unsigned+int+flags)+{
++++return+bool(value+&+flags);
}

inline+bool+all_flags_present(unsigned+int+value,+unsigned+int+flags)+{
++++return+(value+&+flags)+==+flags;
}

enum:+unsigned+int+{
++++ERROR_1+=+1U,
++++ERROR_2+=+2U,+//+or+0b10
++++ERROR_3+=+4U,+//+or+0b100
++++SUCCESS_1+=+8U,
++++SUCCESS_2+=+16U,
++++OTHER_FLAG+=+32U,
};

int+main(void)
{
++++unsigned+int+value+=+0b101011;+//+ERROR_1,+ERROR_2,+SUCCESS_1,+OTHER_FLAG
++++unsigned+int+all_error_flags+=+ERROR_1+%7C+ERROR_2+%7C+ERROR_3;
++++unsigned+int+all_success_flags+=+SUCCESS_1+%7C+SUCCESS_2;
++++
++++std::cout+<<+"Was+there+at+least+one+error:+"+<<+any_flag_present(value,+all_error_flags)+<<+std::endl;
++++std::cout+<<+"Are+all+success+flags+enabled:+"+<<+all_flags_present(value,+all_success_flags)+<<+std::endl;
++++std::cout+<<+"Is+the+other+flag+enabled+with+eror+1:+"+<<+all_flags_present(value,+ERROR_1+%7C+OTHER_FLAG)+<<+std::endl;
++++return+0;
}|code-block|syntax|javascript|309790|entityMap|0|LINK|mutability|MUTABLE|url|https://rextester.com/XIKE82408^0|0|0|0|C|V|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|Z|8|@]|9|@$H|10|I|11|1|12]]|A|$]]|$1|J|3|K|5|L|7|13|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

The precedent answers show you how to handle bit checks, but more often then not, it is all about flags encoded in an integer, which is not well defined in any of the precedent cases.
In a typical scenario, flags are defined as integers themselves, with a bit to 1 for the specific bit it refers to. In the example hereafter, you can check if the integer has ANY flag from a list of flags (multiple error flags concatenated) or if EVERY flag is in the integer (multiple success flags concatenated).
Following an example of how to handle flags in an integer.
Live example available here:
<a href="https://rextester.com/XIKE82408" rel="nofollow noreferrer">https://rextester.com/XIKE82408</a>
<pre class="lang-cpp prettyprint-override"><code>//g++ 7.4.0

#include &lt;iostream&gt;
#include &lt;stdint.h&gt;

inline bool any_flag_present(unsigned int value, unsigned int flags) {
 return bool(value &amp; flags);
}

inline bool all_flags_present(unsigned int value, unsigned int flags) {
 return (value &amp; flags) == flags;
}

enum: unsigned int {
 ERROR_1 = 1U,
 ERROR_2 = 2U, // or 0b10
 ERROR_3 = 4U, // or 0b100
 SUCCESS_1 = 8U,
 SUCCESS_2 = 16U,
 OTHER_FLAG = 32U,
};

int main(void)
{
 unsigned int value = 0b101011; // ERROR_1, ERROR_2, SUCCESS_1, OTHER_FLAG
 unsigned int all_error_flags = ERROR_1 | ERROR_2 | ERROR_3;
 unsigned int all_success_flags = SUCCESS_1 | SUCCESS_2;
 
 std::cout &lt;&lt; &quot;Was there at least one error: &quot; &lt;&lt; any_flag_present(value, all_error_flags) &lt;&lt; std::endl;
 std::cout &lt;&lt; &quot;Are all success flags enabled: &quot; &lt;&lt; all_flags_present(value, all_success_flags) &lt;&lt; std::endl;
 std::cout &lt;&lt; &quot;Is the other flag enabled with eror 1: &quot; &lt;&lt; all_flags_present(value, ERROR_1 | OTHER_FLAG) &lt;&lt; std::endl;
 return 0;
}
</code></pre>

blocks|key|4374823|text|我这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4374824|LATGbits.LATG0=((m&0x8)>0)；//检查m的bit-2是否为1|4374825|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

I make this:

LATGbits.LATG0=((m&amp;0x8)>0); //to check if bit-2 of m is 1

blocks|key|3527514|text|最快最好的宏|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3527515|#define+get_bit_status()++++(+YOUR_VAR++&+++(+1+<<+BITX+)+)

.
.

if+(get_rx_pin_status()+==+(+1+<<+BITX+))
{
++++do();
}|code-block|syntax|javascript|3527516|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|$]]

fast and best macro
<pre><code>#define get_bit_status() ( YOUR_VAR &amp; ( 1 &lt;&lt; BITX ) )

.
.

if (get_rx_pin_status() == ( 1 &lt;&lt; BITX ))
{
 do();
}
</code></pre>

blocks|key|4374590|text|最快的方法似乎是查找掩码的表。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4374591|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|$]]

the fastest way seems to be a lookup table for masks

<pre><code>int temp = 0x5E; // in binary 0b1011110.
</code></pre>

Is there such a way to check if bit 3 in temp is 1 or 0 without bit shifting and masking.

Just want to know if there is some built in function for this, or am I forced to write one myself.

C/C++ check if one bit is set in, i.e. int variable

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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int temp = 0x5E; // in binary 0b1011110.有没有这样一种方法来检查temp中的位3是1还是0，而不需要位移位和掩码。我只想知道是否有一些内置的函数，或者我被迫自己写一个。

问C/C++检查是否设置了一位，即int变量
EN

回答 22

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问C/C++检查是否设置了一位，即int变量EN

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问C/C++检查是否设置了一位，即int变量
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