blocks|key|3390629|text|根据6.3.3.3中的C90标准：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3390630|3390631|一元%2B运算符的结果是其操作数的值。在操作数上执行积分提升。并且结果具有提升类型。|blockquote|offset|length|style|BOLD|3390632|3390633|3390634|和|3390635|3390636|一元%2B或-运算符的操作数应为算术类型..|3390637|3390638|entityMap^0|0|0|F|P|0|0|0|0|1|0|0|0|K|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|U|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|V|8|@$F|W|G|X|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|10|8|@$F|11|G|12|H|I]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|P|5|E|7|14|8|@$F|15|G|16|H|I]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|S|$]]

As per the C90 standard in 6.3.3.3:

<blockquote>
 The result of the unary + operator is the value of its operand. The integral promotion is 
 performed on the operand. and the result has the promoted type. 
</blockquote>

and 

<blockquote>
 The operand of the unary + or - operator shall have arithmetic type.. 
</blockquote>

blocks|key|3105599|text|您可以将其用作表达式具有算术类型的一种断言：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3105600|#define+CHECK_ARITHMETIC(x)+(%2B(x))|code-block|syntax|javascript|3105601|如果x的计算结果为(比方说)一个指针，这将生成一个编译时错误。|offset|length|style|CODE|3105602|这大概是我能想到的唯一实际用途了。|3105603|entityMap^0|0|0|2|1|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|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|$]]

You can use it as a sort of assertion that an expression has arithmetic type:

<pre><code>#define CHECK_ARITHMETIC(x) (+(x))
</code></pre>

This will generate a compile-time error if <code>x</code> evaluates to (say) a pointer.

That is about the only practical use I can think of.

blocks|key|3390679|text|据我所知，一元加号运算符有一个非常方便的用法:在宏中。假设你想做一些类似这样的事情|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3390680|#if+FOO+>+0|code-block|syntax|javascript|3390681|如果FOO未定义，在这种情况下，C语言要求将其替换为0。但是如果FOO的定义是空的，那么上面的指令将导致错误。相反，您可以使用：|offset|length|style|CODE|3390682|#if+FOO%2B0+>+0|3390683|现在，无论FOO是未定义的、定义为空的还是定义为整数值的，指令在语法上都是正确的。|3390684|当然，这是否会产生所需的语义是一个完全不同的问题，但在某些有用的情况下，它会。|3390685|编辑：请注意，您甚至可以使用它来区分FOO定义为零与定义为空的情况，如下所示：|BOLD|3390686|#if+2*FOO%2B1+==+1
/*+FOO+is+0+*/
#else
/*+FOO+is+blank+*/
#endif|3390687|entityMap^0|0|0|2|3|W|3|0|0|5|3|0|0|0|3|I|3|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|11|8|@$I|12|J|13|K|L]|$I|14|J|15|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|17|8|@$I|18|J|19|K|L]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1A|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1B|8|@$I|1C|J|1D|K|U]|$I|1E|J|1F|K|L]]|9|@]|A|$]]|$1|V|3|W|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|X|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|Y|$]]

There's one very handy use of the unary plus operator I know of: in macros. Suppose you want to do something like

<pre><code>#if FOO &gt; 0
</code></pre>

If <code>FOO</code> is undefined, the C language requires it be replaced by 0 in this case. But if <code>FOO</code> was defined with an empty definition, the above directive will result in an error. Instead you can use:

<pre><code>#if FOO+0 &gt; 0
</code></pre>

And now, the directive will be syntactically correct whether <code>FOO</code> is undefined, defined as blank, or defined as an integer value.

Of course whether this will yield the desired semantics is a completely separate question, but in some useful cases it will.

Edit: Note that you can even use this to distinguish the cases of <code>FOO</code> being defined as zero versus defined as blank, as in:

<pre><code>#if 2*FOO+1 == 1
/* FOO is 0 */
#else
/* FOO is blank */
#endif
</code></pre>

blocks|key|3390563|text|差不多吧。它主要是为了完整性，并使这样的结构看起来更干净：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3390564|int+arr[]+=+{
++++%2B4,
++++-1,
++++%2B1,
++++-4,
};|code-block|syntax|javascript|3390565|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|$]]

Pretty much. It's mainly present for completeness, and to make constructions like this look a little cleaner:

<pre><code>int arr[] = {
 +4,
 -1,
 +1,
 -4,
};
</code></pre>

blocks|key|3390766|text|我发现一元%2B运算符做两件事|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3390767|3390768|integer+promotion|unordered-list-item|3390769|turning+lvalue+into+rvalue|3390770|3390771|整数提升示例：|BOLD|3390772|#include+<stdio.h>

int+main(void)+{

++++char+ch;
++++short+sh;
++++int+i;
++++long+l;

++++printf("%25d+%25d+%25d+%25d\n",sizeof(ch),sizeof(sh),sizeof(i),sizeof(l));
++++printf("%25d+%25d+%25d+%25d\n",sizeof(%2Bch),sizeof(%2Bsh),sizeof(%2Bi),sizeof(%2Bl));
++++return+0;
}|code-block|syntax|javascript|3390773|典型输出(在64位平台上)：|3390774|1+2+4+8
4+4+4+8|3390775|将左值转换为右值的示例：|3390776|int+i=0,j;

j=(%2Bi)%2B%2B;+//+error+lvalue+required|3390777|entityMap^0|5|1|0|0|0|H|0|0|Q|0|0|0|7|0|0|0|0|9|3|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|17|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|J|3|K|5|I|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|L|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|1F|8|@$9|1G|A|1H|B|O]]|D|@]|E|$]]|$1|P|3|Q|5|R|7|1I|8|@]|D|@]|E|$S|T]]|$1|U|3|V|5|6|7|1J|8|@]|D|@]|E|$]]|$1|W|3|X|5|R|7|1K|8|@]|D|@]|E|$S|T]]|$1|Y|3|Z|5|6|7|1L|8|@$9|1M|A|1N|B|O]]|D|@]|E|$]]|$1|10|3|11|5|R|7|1O|8|@]|D|@]|E|$S|T]]|$1|12|3|-4|5|6|7|1P|8|@]|D|@]|E|$]]]|13|$]]

I found two things that unary <code>+</code> operator do is

<ul>
<li><code>integer promotion</code></li>
<li><code>turning lvalue into rvalue</code></li>
</ul>

integer promotion example:

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

int main(void) {

 char ch;
 short sh;
 int i;
 long l;

 printf("%d %d %d %d\n",sizeof(ch),sizeof(sh),sizeof(i),sizeof(l));
 printf("%d %d %d %d\n",sizeof(+ch),sizeof(+sh),sizeof(+i),sizeof(+l));
 return 0;
}
</code></pre>

Typical output (on 64-bit platform):

<pre><code>1 2 4 8
4 4 4 8
</code></pre>

turning lvalue into rvalue example:

<pre><code>int i=0,j;

j=(+i)++; // error lvalue required
</code></pre>

blocks|key|3390730|text|并不完全是无人操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3390731|一元%2B操作符只做一件事:它应用整数提升。因为如果在表达式中使用操作数，这些操作无论如何都会发生，所以可以想象在C中使用一元%2B只是为了与一元-对称。|offset|length|style|CODE|3390732|很难在行动中看到这一点，因为促销活动是如此普遍。|3390733|我想出了这个：|3390734|printf("%25zd\n",+sizeof(+(char)+'x'));
printf("%25zd\n",+sizeof(%2B(char)+'x'));|code-block|syntax|javascript|3390735|它(在我的Mac上)打印|3390736|1
4|3390737|entityMap^0|0|2|1|1P|1|1X|1|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|14|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|16|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|17|8|@]|9|@]|A|$]]|$1|S|3|T|5|N|7|18|8|@]|9|@]|A|$O|P]]|$1|U|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|V|$]]

<h3>Not precisely a no-op</h3>

The unary <code>+</code> operator does only one thing: it applies the integer promotions. Since those would occur anyway if the operand were used in an expression, one imagines that unary <code>+</code> is in C simply for symmetry with unary <code>-</code>.

It's difficult to see this in action because the promotions are so generally applied.

I came up with this:

<pre><code>printf("%zd\n", sizeof( (char) 'x'));
printf("%zd\n", sizeof(+(char) 'x'));
</code></pre>

which (on my Mac) prints

<pre><code>1
4
</code></pre>

blocks|key|1490229|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1490230|+C中一元'%2B‘运算符的用途是什么？|blockquote|1490231|1490232|1490233|为了对称，一元加被添加到C中，而一元减则来自+Rationale+for+International+Standard—Programming+Languages—C|offset|length|1490234|1490235|一元加被C89委员会从几个实现中采用，以实现与一元减的对称。|1490236|1490237|1490238|它不是no-op，它在其操作数上执行整数提升。引用我对Does+Unary+%2B+operator+do+type+conversions?的回答|1490239|草案C99标准部分6.5.3.3一元算术运算符说：|style|CODE|1490240|1490241|一元%2B运算符的结果是其(提升的)操作数的值。在操作数上执行整数提升，并且结果具有提升类型。|BOLD|1490242|1490243|1490244|值得一提的是，Annotated+C%2B%2B+Reference+Manual()提供了以下关于一元加的评论：|1490245|1490246|一元加是历史上的偶然事件，通常是无用的。|1490247|1490248|entityMap|0|LINK|mutability|MUTABLE|url|http://www.open-std.org/jtc1/sc22/wg14/www/C99RationaleV5.10.pdf|1|https://stackoverflow.com/q/24805580/1708801|2|http://www.stroustrup.com/arm.html^0|0|0|0|0|M|1P|0|0|0|0|0|0|R|16|1|0|9|7|0|0|M|4|0|0|0|7|W|2|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|1I|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1J|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|1K|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|1L|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|1M|7|@]|8|@$H|1N|I|1O|1|1P]]|9|$]]|$1|J|3|-4|4|5|6|1Q|7|@]|8|@]|9|$]]|$1|K|3|L|4|C|6|1R|7|@]|8|@]|9|$]]|$1|M|3|-4|4|5|6|1S|7|@]|8|@]|9|$]]|$1|N|3|-4|4|5|6|1T|7|@]|8|@]|9|$]]|$1|O|3|P|4|5|6|1U|7|@]|8|@$H|1V|I|1W|1|1X]]|9|$]]|$1|Q|3|R|4|5|6|1Y|7|@$H|1Z|I|20|S|T]]|8|@]|9|$]]|$1|U|3|-4|4|5|6|21|7|@]|8|@]|9|$]]|$1|V|3|W|4|C|6|22|7|@$H|23|I|24|S|X]]|8|@]|9|$]]|$1|Y|3|-4|4|5|6|25|7|@]|8|@]|9|$]]|$1|Z|3|-4|4|5|6|26|7|@]|8|@]|9|$]]|$1|10|3|11|4|5|6|27|7|@]|8|@$H|28|I|29|1|2A]]|9|$]]|$1|12|3|-4|4|5|6|2B|7|@]|8|@]|9|$]]|$1|13|3|14|4|C|6|2C|7|@]|8|@]|9|$]]|$1|15|3|-4|4|5|6|2D|7|@]|8|@]|9|$]]|$1|16|3|-4|4|5|6|2E|7|@]|8|@]|9|$]]]|17|$18|$4|19|1A|1B|9|$1C|1D]]|1E|$4|19|1A|1B|9|$1C|1F]]|1G|$4|19|1A|1B|9|$1C|1H]]]]

<blockquote>
 What is the purpose of the unary '+' operator in C?
</blockquote>

Unary plus was added to C for symmetry with unary minus, from the <a href="http://www.open-std.org/jtc1/sc22/wg14/www/C99RationaleV5.10.pdf" rel="noreferrer"> Rationale for International Standard—Programming Languages—C</a>:

<blockquote>
 Unary plus was adopted by the C89 Committee from several implementations, for symmetry with unary minus.
</blockquote>

and it is not a no-op, it performs the integer promotions on its operand. Quoting from my answer to <a href="https://stackoverflow.com/q/24805580/1708801">Does Unary + operator do type conversions?</a>:

The draft C99 standard section <code>6.5.3.3</code> Unary arithmetic operators says:

<blockquote>
 The result of the unary + operator is the value of its (promoted)
 operand. The integer promotions are performed on the operand, and the
 result has the promoted type.
</blockquote>

Worth pointing out that <a href="http://www.stroustrup.com/arm.html" rel="noreferrer">Annotated C++ Reference Manual(ARM)</a> provides the following commentary on unary plus:

<blockquote>
 Unary plus is a historical accident and generally useless.
</blockquote>

blocks|key|3390704|text|你说的“no-op”是指汇编指令吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3390705|如果是，那肯定不是。|3390706|%2B4只有4个-编译器不会再添加任何指令。|offset|length|style|BOLD|3390707|entityMap^0|0|0|0|6|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|N|8|@$F|O|G|P|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

By 'no-op', do you mean the assembly instruction? 
If so, then definitely not.

+4 is just 4 - the compiler won't add any further instructions.

In C, it's legal to write something like:

<pre><code>int foo = +4;
</code></pre>

However, as far as I can tell, the unary plus (<code>+</code>) in <code>+4</code> is a no-op. Is it?

What is the purpose of the unary plus (+) operator 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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在C语言中，可以合法地编写如下代码：int foo = +4;然而，据我所知，+4中的一元加号(+)是没有意义的。是吗?

问在C中一元加号(+)运算符的用途是什么？
EN

回答 8

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问在C中一元加号(+)运算符的用途是什么？
EN