blocks|key|1402325|text|这里有点混乱，让我们把它弄清楚。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1402326|0b1010是一个整数字面值，是一个用基2写的常数，编译时整数值。同样，0xA在基16中是文字，10在基10中。所有这些都是指相同的整数，这只是告诉编译器您指的是哪个数字的另一种方式。在运行时，在内存中，这个整数总是表示为一个基-2数字.|unordered-list-item|offset|length|style|CODE|BOLD|1402327|std::cout+<<+a；获取a的整数值，并输出其字符串表示形式。默认情况下，它在基10中输出它，但是您可以使用std::hex修饰符让它在基16中输出。没有预定义的修饰符要在二进制中打印。所以你需要自己去做(或者用谷歌搜索，这是一个常见的问题)。|1402328|最后，0b只用于定义整数字面值。它是而不是运行时操作符。回想一下，所有的ints都表示为内存中的基2数字。从机器的角度来看，其他基础不存在，int是int，所以没有什么可转换的。如果需要从字符串中读取二进制数字，则需要将反向代码滚到打印它的操作(std::cin+>>+n假设输入是一个基数10的数字，因此如果输入实际上位于基2中，则读取错误的数字)。|1402329|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/ios/hex/^0|0|0|6|10|3|1C|2|1K|C|0|0|E|H|1|1M|8|27|E|1M|8|0|0|3|2|10|3|1Y|3|22|3|3F|D|I|3|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@$E|X|F|Y|G|H]|$E|Z|F|10|G|H]|$E|11|F|12|G|H]|$E|13|F|14|G|I]]|9|@]|A|$]]|$1|J|3|K|5|D|7|15|8|@$E|16|F|17|G|H]|$E|18|F|19|G|H]|$E|1A|F|1B|G|H]|$E|1C|F|1D|G|I]]|9|@$E|1E|F|1F|1|1G]]|A|$]]|$1|L|3|M|5|D|7|1H|8|@$E|1I|F|1J|G|H]|$E|1K|F|1L|G|H]|$E|1M|F|1N|G|H]|$E|1O|F|1P|G|H]|$E|1Q|F|1R|G|H]|$E|1S|F|1T|G|I]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]]|O|$P|$5|Q|R|S|A|$T|U]]]]

There is a bit of confusion here, let's disentangle it a bit.

<ul>
<li><code>0b1010</code> is an integer literal, a constant, compile-time integer value written in base 2. Likewise, <code>0xA</code> is a literal in base 16 and <code>10</code> is in base 10. All of these refer to the same integer, it is just a different way of telling the compiler which number you mean. At runtime, in memory, this integer is always represented as a base-2 number.</li>
<li><code>std::cout &lt;&lt; a</code>; takes the integer value of <code>a</code> and outputs a string representation of it. By default it outputs it in base 10, but you can i.e use the <a href="http://www.cplusplus.com/reference/ios/hex/" rel="nofollow"><code>std::hex</code></a> modifier to have it output it in base 16. There is no predefined modifier to print in binary. So you need to do that on your own (or google it, it is a common question).</li>
<li><code>0b</code> at last, is only used to define integer literals. It is not a runtime operator. Recall, all <code>int</code>s are represented as base 2 numbers in memory. Other bases do not exist from a machine point of view, <code>int</code> is <code>int</code>, so there is nothing to convert. If you need to read a binary number from a string, you would roll the reverse code to what you do to print it (<code>std::cin &gt;&gt; n</code> assumes that the input is a base 10 number, so it reads a wrong number if the input is actually intended to be in base 2).</li>
</ul>

blocks|key|164637|text|虽然没有直接读取二进制数字的函数，但是有一些函数，strtox+(其中x表示数据类型)将包含二进制数字(或任何其他基数)的字符串转换为数值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|164638|因此，解决方案是首先将数字读入字符串，然后转换它。|164639|示例：|164640|char+input[100];
char+*endpointer;

<read+input+using+either+C+or+C%2B%2B+syntax>

int+n+=+(int)+strtol(input,+&endpointer,+2);|code-block|syntax|javascript|164641|entityMap^0|P|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|U|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|V|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

While there is no function to read binary numbers directly, there are functions, <code>strtox</code> (where x represents the data type) to convert a string containing a binary number (or a number of any other base) to a numeric value.

So the solution is to first read the number as a string and then convert it.

Example:

<pre><code>char input[100];
char *endpointer;

&lt;read input using either C or C++ syntax&gt;

int n = (int) strtol(input, &amp;endpointer, 2);
</code></pre>

blocks|key|539643|text|不如你自己做吧：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|539644|uint32_t+a+=+0;

char+c;
while+((c+=+getchar())+!=+'\n')+{+//+read+a+line+char+by+char
++a+<<=+1;++++++++++++++++++++++++//+shift+the+uint32+a+bit+left
++a+%2B=+(c+-+'0')+&+1;+++++++++++++//+convert+the+char+to+0/1+and+put+it+at+the+end+of+the+binary
}

printf("%25u\n",+a);|code-block|syntax|javascript|539645|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|$]]

rather do it yourself: 

<pre><code>uint32_t a = 0;

char c;
while ((c = getchar()) != '\n') { // read a line char by char
 a &lt;&lt;= 1; // shift the uint32 a bit left
 a += (c - '0') &amp; 1; // convert the char to 0/1 and put it at the end of the binary
}

printf("%u\n", a);
</code></pre>

blocks|key|539001|text|要将二进制数字作为输入，我经常使用两种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|539002|(基调：以输入为字符串！)用途：#include+<string>)|offset|length|style|BOLD|CODE|539003|位集库bitset模板的to_ulong()方法|ordered-list-item|539004|为此，需要包括使用bitset的#include+<bitset>库。|unordered-list-item|539005|539006|示例：|539007|string+s;
cin>>s;+//+Suppose+s+=+"100100101"
int+n+=+(int)+bitset<64>(s).to_ulong();
cout<<n;+//+293|code-block|syntax|javascript|539008|探索更多关于bitset+这里和to_ulong()+这里的信息。|blockquote|539009|string库的stoi()方法|539010|为此，需要包括使用string的#include+<string>库。|539011|539012|539013|string+s;
cin>>s;+//+Suppose+s+=+"100100101"
int+n+=+stoi(s,+0,+2);
cout<<n;+//+293|539014|探索stoi()+这里的格式。|539015|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/bitset/bitset/|1|https://peter.bloomfield.online/convert-a-number-to-a-binary-string-and-back-in-cpp/|2|http://www.cplusplus.com/reference/string/stoi/^0|0|1|2|G|H|0|3|6|3|6|C|A|1|9|6|9|6|G|H|1|0|0|3|0|0|6|6|G|A|D|2|0|R|2|1|0|0|6|0|6|8|6|1|9|6|9|6|G|H|1|0|0|3|0|0|2|6|9|2|2|0^^$0|@$1|2|3|4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1M|8|@$D|1N|E|1O|F|G]|$D|1P|E|1Q|F|H]]|9|@]|A|$]]|$1|I|3|J|5|K|7|1R|8|@$D|1S|E|1T|F|G]|$D|1U|E|1V|F|H]|$D|1W|E|1X|F|H]]|9|@]|A|$]]|$1|L|3|M|5|N|7|1Y|8|@$D|1Z|E|20|F|G]|$D|21|E|22|F|H]|$D|23|E|24|F|H]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|25|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|26|8|@$D|27|E|28|F|G]]|9|@]|A|$]]|$1|R|3|S|5|T|7|29|8|@]|9|@]|A|$U|V]]|$1|W|3|X|5|Y|7|2A|8|@$D|2B|E|2C|F|H]|$D|2D|E|2E|F|H]]|9|@$D|2F|E|2G|1|2H]|$D|2I|E|2J|1|2K]]|A|$]]|$1|Z|3|10|5|K|7|2L|8|@$D|2M|E|2N|F|G]|$D|2O|E|2P|F|H]|$D|2Q|E|2R|F|H]]|9|@]|A|$]]|$1|11|3|12|5|N|7|2S|8|@$D|2T|E|2U|F|G]|$D|2V|E|2W|F|H]|$D|2X|E|2Y|F|H]]|9|@]|A|$]]|$1|13|3|-4|5|6|7|2Z|8|@]|9|@]|A|$]]|$1|14|3|Q|5|6|7|30|8|@$D|31|E|32|F|G]]|9|@]|A|$]]|$1|15|3|16|5|T|7|33|8|@]|9|@]|A|$U|V]]|$1|17|3|18|5|Y|7|34|8|@$D|35|E|36|F|H]]|9|@$D|37|E|38|1|39]]|A|$]]|$1|19|3|-4|5|6|7|3A|8|@]|9|@]|A|$]]]|1A|$1B|$5|1C|1D|1E|A|$1F|1G]]|1H|$5|1C|1D|1E|A|$1F|1I]]|1J|$5|1C|1D|1E|A|$1F|1K]]]]

To take a binary number as input, there are two ways I use frequently:
(Keynote: Take the input as string!!! use: <code>#include &lt;string&gt;</code>)
<ol>
<li>The <code>to_ulong()</code> method of the <code>bitset</code> template of the bitset library
<ul>
<li>for this you need to include the <code>bitset</code> library using <code>#include &lt;bitset&gt;</code></li>
</ul>
</li>
</ol>
Example:
<pre><code>string s;
cin&gt;&gt;s; // Suppose s = &quot;100100101&quot;
int n = (int) bitset&lt;64&gt;(s).to_ulong();
cout&lt;&lt;n; // 293
</code></pre>
<blockquote>
Explore more about <code>bitset</code> <a href="http://www.cplusplus.com/reference/bitset/bitset/" rel="nofollow noreferrer">here</a> and about <code>to_ulong()</code> <a href="https://peter.bloomfield.online/convert-a-number-to-a-binary-string-and-back-in-cpp/" rel="nofollow noreferrer">here</a>.
</blockquote>
<ol start="2">
<li>The <code>stoi()</code> method of the <code>string</code> library
<ul>
<li>for this you need to include the <code>string</code> library using <code>#include &lt;string&gt;</code></li>
</ul>
</li>
</ol>
Example:
<pre><code>string s;
cin&gt;&gt;s; // Suppose s = &quot;100100101&quot;
int n = stoi(s, 0, 2);
cout&lt;&lt;n; // 293
</code></pre>
<blockquote>
Explore the format of <code>stoi()</code> <a href="http://www.cplusplus.com/reference/string/stoi/" rel="nofollow noreferrer">here</a>.
</blockquote>

Is there a way for the user to input a binary number in C or C++?

If we write something like

<pre><code>int a = 0b1010;
std::cout &lt;&lt; a &lt;&lt; std::endl
</code></pre>

Then the output comes out to be 10 (when using the appropriate compiler extensions).

but when we try to write

<pre><code>int n;
std::cin &gt;&gt; n;
int t = 0bn;
</code></pre>

It gives us an error so can anyone suggest that how can we directly read binary number as input rather than using string to store input?

How to read a binary number as input?

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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用户是否有办法在C或C++中输入二进制数字？如果我们写这样的东西int a = 0b1010;std::cout << a << std::endl然后输出为10 (当使用适当的编译器扩展时)。但当我们试着写int n;std::cin >> n;int t = 0bn;它给了我们一个错误，所以有人能建议我们如何直接读...

问如何读取二进制数字作为输入？
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问如何读取二进制数字作为输入？EN