blocks|key|1686441|text|它实际上并不打印空白，但很可能是值为5的ASCII字符，它是不可打印的(或不可见的)。有许多invisible+ASCII+character+codes，其中大多数低于值32，这实际上是空白。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1686442|您必须将aa转换为unsigned+int才能输出数值，因为ostream&+operator<<(ostream&,+unsigned+char)会尝试输出可见字符值。|style|CODE|1686443|uint8_t+aa=5;

cout+<<+"value+is+"+<<+unsigned(aa)+<<+endl;|code-block|syntax|javascript|1686444|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cs.tut.fi/~jkorpela/chars/c0.html^0|1A|V|0|0|4|2|9|C|U|18|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@$A|Z|B|10|F|G]|$A|11|B|12|F|G]|$A|13|B|14|F|G]]|9|@]|C|$]]|$1|H|3|I|5|J|7|15|8|@]|9|@]|C|$K|L]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

It doesn't really print a blank, but most probably the ASCII character with value 5, which is non-printable (or invisible). There's a number of <a href="http://www.cs.tut.fi/~jkorpela/chars/c0.html" rel="noreferrer">invisible ASCII character codes</a>, most of them below value 32, which is the blank actually.

You have to convert <code>aa</code> to <code>unsigned int</code> to output the numeric value, since <code>ostream&amp; operator&lt;&lt;(ostream&amp;, unsigned char)</code> tries to output the visible character value.

<pre><code>uint8_t aa=5;

cout &lt;&lt; "value is " &lt;&lt; unsigned(aa) &lt;&lt; endl;
</code></pre>

blocks|key|1686513|text|uint8_t很可能是unsigned+char的typedef。ostream类对unsigned+char有一个特殊的重载，即它打印数字为5的字符，这是不可打印的，因此是空格。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1686514|entityMap^0|0|7|B|D|P|7|X|7|16|D|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]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

<code>uint8_t</code> will most likely be a <code>typedef</code> for <code>unsigned char</code>. The <code>ostream</code> class has a special overload for <code>unsigned char</code>, i.e. it prints the character with the number 5, which is non-printable, hence the empty space.

blocks|key|1900258|text|在任何原始数据类型的变量前添加一个一元%2B运算符将产生可打印的数值，而不是ASCII字符(对于char类型)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1900259|uint8_t+aa+=+5;
cout<<"value+is+"<<+%2Baa+<<endl;+//+value+is+5|code-block|syntax|javascript|1900260|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|$]]

Adding a unary + operator before the variable of any primitive data type will give printable numerical value instead of ASCII character(in case of char type).
<pre><code>uint8_t aa = 5;
cout&lt;&lt;&quot;value is &quot;&lt;&lt; +aa &lt;&lt;endl; // value is 5
</code></pre>

blocks|key|1900042|text|这是因为输出操作符将uint8_t视为char+(uint8_t通常只是unsigned+char的别名)，所以它使用ASCII码(这是最常见的字符编码系统)+5打印字符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1900043|例如，参见this+reference。|1900044|entityMap|0|LINK|mutability|MUTABLE|url|http://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt2^0|A|7|J|4|P|7|10|D|28|1|0|5|E|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]]|E|$]]|$1|H|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

It's because the output operator treats the <code>uint8_t</code> like a <code>char</code> (<code>uint8_t</code> is usually just an alias for <code>unsigned char</code>), so it prints the character with the ASCII code (which is the most common character encoding system) <code>5</code>.

See e.g. <a href="http://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt2" rel="noreferrer">this reference</a>.

blocks|key|1900112|text|cout正在将aa视为ASCII值5的char，这是一个无法打印的字符，请尝试在打印前将其类型转换为int。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1900113|entityMap^0|0|4|7|2|H|1|J|4|1E|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]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

<code>cout</code> is treating <code>aa</code> as <code>char</code> of ASCII value <code>5</code> which is an unprintable character, try typecasting to <code>int</code> before printing.

blocks|key|1900289|text|std::ostream和char之间的operator<<()重载是一个非成员函数。您可以显式地使用成员函数将char+(或uint8_t)视为int。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1900290|#include+<iostream>
#include+<cstddef>

int+main()
{
+++uint8_t+aa=5;

+++std::cout+<<+"value+is+";
+++std::cout.operator<<(aa);
+++std::cout+<<+std::endl;

+++return+0;
}|code-block|syntax|javascript|1900291|输出：|1900292|value+is+5|1900293|entityMap^0|0|C|D|4|K|C|1K|4|1R|7|21|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|P|$]]

The <code>operator&lt;&lt;()</code> overload between <code>std::ostream</code> and <code>char</code> is a non-member function. You can explicitly use the member function to treat a <code>char</code> (or a <code>uint8_t</code>) as an <code>int</code>.
<pre><code>#include &lt;iostream&gt;
#include &lt;cstddef&gt;

int main()
{
 uint8_t aa=5;

 std::cout &lt;&lt; &quot;value is &quot;;
 std::cout.operator&lt;&lt;(aa);
 std::cout &lt;&lt; std::endl;

 return 0;
}
</code></pre>
Output:
<pre><code>value is 5
</code></pre>

blocks|key|1686715|text|正如其他人之前所说的，之所以会出现这个问题，是因为标准流将有符号字符和无符号字符视为单个字符而不是数字。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1686716|以下是我的解决方案，只需最少的代码更改：|1686717|uint8_t+aa+=+5;

cout+<<+"value+is+"+<<+aa+%2B+0+<<+endl;|code-block|syntax|javascript|1686718|对于包括浮点在内的任何数字，添加"%2B0"都是安全的。|offset|length|style|CODE|1686719|对于整数类型，如果为sizeof(aa)+<+sizeof(int)，它会将结果类型更改为int。如果为sizeof(aa)+>=+sizeof(int)，则不会更改类型。|1686720|这个解决方案对于准备要打印成流的int8_t也很好，而其他一些解决方案就不那么好了：|1686721|int8_t+aa+=+-120;

cout+<<+"value+is+"+<<+aa+%2B+0+<<+endl;
cout+<<+"bad+value+is+"+<<+unsigned(aa)+<<+endl;|1686722|输出：|1686723|value+is+-120
bad+value+is+4294967176|1686724|附注:由pepper_chico和πάνταῥεῖ提供的ADL解决方案非常漂亮。|1686725|entityMap^0|0|0|0|G|4|0|A|O|19|3|1G|P|0|G|6|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|15|8|@$K|16|L|17|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|18|8|@$K|19|L|1A|M|N]|$K|1B|L|1C|M|N]|$K|1D|L|1E|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1F|8|@$K|1G|L|1H|M|N]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1J|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|Y|3|Z|5|6|7|1L|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]]|11|$]]

As others said before the problem occurs because standard stream treats signed char and unsigned char as single characters and not as numbers.

Here is my solution with minimal code changes:

<pre><code>uint8_t aa = 5;

cout &lt;&lt; "value is " &lt;&lt; aa + 0 &lt;&lt; endl;
</code></pre>

Adding <code>"+0"</code> is safe with any number including floating point.

For integer types it will change type of result to <code>int</code> if <code>sizeof(aa) &lt; sizeof(int)</code>. And it will not change type if <code>sizeof(aa) &gt;= sizeof(int)</code>.

This solution is also good for preparing <code>int8_t</code> to be printed to stream while some other solutions are not so good:

<pre><code>int8_t aa = -120;

cout &lt;&lt; "value is " &lt;&lt; aa + 0 &lt;&lt; endl;
cout &lt;&lt; "bad value is " &lt;&lt; unsigned(aa) &lt;&lt; endl;
</code></pre>

Output:

<pre><code>value is -120
bad value is 4294967176
</code></pre>

P.S. Solution with ADL given by pepper_chico and πάντα ῥεῖ is really beautiful.

I have a weird problem about working with integers in C++.

I wrote a simple program that sets a value to a variable and then prints it, but it is not working as expected.

My program has only two lines of code:

<pre><code>uint8_t aa = 5;

cout &lt;&lt; "value is " &lt;&lt; aa &lt;&lt; endl;
</code></pre>

The output of this program is <code>value is</code>

I.e., it prints blank for <code>aa</code>.

When I change <code>uint8_t</code> to <code>uint16_t</code> the above code works like a charm.

I use Ubuntu&nbsp;12.04 (Precise Pangolin), 64-bit, and my compiler version is:

<pre><code>gcc version 4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5)
</code></pre>

uint8_t can't be printed with cout

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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我有一个关于在C++中处理整数的奇怪问题。我编写了一个简单的程序，将一个值设置为一个变量，然后打印出来，但它并没有像预期的那样工作。我的程序只有两行代码：uint8_t aa = 5;cout << "value is " << aa << endl;这个程序的输出是value is也就是说，它为aa打印空白。当我将uint8_t更改为uint16_t时，上面的代码非常出色。我使用的是Ubuntu

问无法使用cout打印uint8_t
EN

回答 7

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问无法使用cout打印uint8_t
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