blocks|key|1895406|text|Edit2显然，我之前的建议是改变|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1895407|const+char*+a+=+"asdfgh"|code-block|syntax|javascript|1895408|至|1895409|const+char[6]+a=+"asdfgh"|1895410|很管用。|1895411|我仍然在研究它，但是当你声明一个指针时，你只是说这个指针是常量，而不是它所指向的内存。当你声明一个数组时，你是在说const后面的数据也是不可变的。我正在进一步调查。|1895412|Edit3：this的第二段解释了发生了什么。|1895413|此外，“干杯”的答案很好地解释了这一点。|1895414|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/forum/general/4495/0^0|0|5|0|0|0|0|0|0|6|4|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1A|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1B|8|@]|D|@]|E|$]]|$1|S|3|T|5|6|7|1C|8|@]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|U|3|V|5|6|7|1G|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]]]

Edit2 Apparently, my earlier suggestion of changing

<pre><code>const char* a = "asdfgh"
</code></pre>

to

<pre><code>const char[6] a= "asdfgh"
</code></pre>

works. 

I'm still looking into it, but when you declare a pointer, you're just saying that that pointer is constant, not the memory it's pointing to. When you declare an array, you're saying that the data behind the const is immutable as well. I'm looking into it further.

Edit3: The second paragraph of <a href="http://www.cplusplus.com/forum/general/4495/0" rel="nofollow">this</a> explains what's going on.

Also, "Cheers'" answer explains it well.

blocks|key|5323492|text|奇伊的答案比我的好，但是...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5323493|尝试将声明更改为：|5323494|extern+const+char+*+const+a;|code-block|syntax|javascript|5323495|然后在其中一个cpp文件中，这样写：|5323496|const+char+*+a+=+"asdf";|5323497|(不过，如果你有两个常量，|5323498|const+char+*+const+a+=+"asdf";|5323499|应该可以工作)。|5323500|除非你同时拥有两个常量，否则你不会得到静态链接。因此，如果同时使用两个const，它应该可以工作，无论是外部的还是静态的。|5323501|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Cheer's answer is better than mine, but...

Try changing your declaration to:

<pre><code>extern const char * const a;
</code></pre>

and then in ONE of your cpp files, say:

<pre><code>const char * a = "asdf";
</code></pre>

(though if you have both consts,

<pre><code>const char * const a = "asdf";
</code></pre>

should work).

You don't get static linkage unless you have both consts. So if you use both consts, it should work, either extern or static.

blocks|key|1283887|text|const+char+a='4';|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1283888|手段|unstyled|1283889|char+const+a+=+'4';|1283890|您可以将其反向读取为“4是a的初始值，它是一个常量char”。|offset|length|style|CODE|1283891|由于此a是恒定的，因此它具有内部链接(即，它不会暴露给其他翻译单元)。|BOLD|1283892|另一方面，|1283893|const+char+*a="asdfgh";|1283894|1283895|char+const+*+a+=+"asdfgh";|1283896|你可以将其反向读取为“”asdfgh“”是用于初始化a的数组，该数组是指向常量char的指针“。|1283897|在这种情况下，a本身不是const，因此默认情况下不会获得内部链接:它具有外部链接。|1283898|当您将头包含在两个或多个转换单元中时，您会得到两个或多个称为a的全局且不同的对象，这违反了C%2B%2B的单一定义规则(通常被称为ODR)。|1283899|一种解决方法是让a本身成为const，|1283900|char+const+*+const+a+=+"asdfgh";|1283901|现在试着倒着读，弄明白它的意思。|1283902|entityMap^0|0|0|0|D|1|P|4|0|3|1|E|4|0|0|0|0|0|Q|1|13|4|0|7|1|C|5|11|4|0|U|1|1P|3|0|8|1|D|5|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|1D|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1E|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|1F|8|@$K|1G|L|1H|M|N]|$K|1I|L|1J|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|1K|8|@$K|1L|L|1M|M|N]|$K|1N|L|1O|M|Q]]|9|@]|A|$]]|$1|R|3|S|5|F|7|1P|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1Q|8|@]|9|@]|A|$B|C]]|$1|V|3|E|5|F|7|1R|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1S|8|@]|9|@]|A|$B|C]]|$1|Y|3|Z|5|F|7|1T|8|@$K|1U|L|1V|M|N]|$K|1W|L|1X|M|N]]|9|@]|A|$]]|$1|10|3|11|5|F|7|1Y|8|@$K|1Z|L|20|M|N]|$K|21|L|22|M|N]|$K|23|L|24|M|Q]]|9|@]|A|$]]|$1|12|3|13|5|F|7|25|8|@$K|26|L|27|M|N]|$K|28|L|29|M|Q]]|9|@]|A|$]]|$1|14|3|15|5|F|7|2A|8|@$K|2B|L|2C|M|N]|$K|2D|L|2E|M|N]]|9|@]|A|$]]|$1|16|3|17|5|6|7|2F|8|@]|9|@]|A|$B|C]]|$1|18|3|19|5|F|7|2G|8|@]|9|@]|A|$]]|$1|1A|3|-4|5|F|7|2H|8|@]|9|@]|A|$]]]|1B|$]]

<pre><code>const char a='4';
</code></pre>

means

<pre><code>char const a = '4';
</code></pre>

which you can read backwards as &ldquo;4 is the initial value for <code>a</code> which is a constant <code>char</code>&rdquo;.

Since this <code>a</code> is constant, it has internal linkage (i.e. it is not exposed to other translation units).

On the other hand,

<pre><code>const char *a="asdfgh";
</code></pre>

means

<pre><code>char const * a = "asdfgh";
</code></pre>

which you can read backwards as &ldquo;"asdfgh" is an array used to initialize <code>a</code> which as a pointer to a constant <code>char</code>&rdquo;.

In this case <code>a</code> itself is not <code>const</code>, and so does not get internal linkage by default: it has external linkage.

When you include your header in two or more translation units, you therefore get two or more global and distinct objects called <code>a</code>, which violates the One Definition Rule of C++ (it&rsquo;s often referred to as just the ODR).

One cure is to make <code>a</code> itself <code>const</code>,

<pre><code>char const * const a = "asdfgh";
</code></pre>

Now try to read that backwards, to make sense of it.

I ran into such problem:

I have one header file:

<pre><code>//first variant: 
#ifndef LIBRARIES_H
#define LIBRARIES_H
 const char a='4'; 
 #endif // LIBRARIES_H
</code></pre>

I include it in few .cpp files.
 Everything works perfect.

But suddenly, if I declare a pointer: 

<pre><code>//second variant:
#ifndef LIBRARIES_H
 #define LIBRARIES_H
 const char *a="asdfgh";
 #endif // LIBRARIES_H
</code></pre>

please, notice,I DECLARE ONLY ONE OF THEM(of variants). 
 I DECLARE a only once(I tried to change name for sadfgh or asdfg).

When I try second variant, I delete first and vise versa.

If I declare a pointer to the string I will get error "multiple inclusion of variable".
 I compile it using qt. I deleted and recompiled project a few times already. 
 <img src="https://i.stack.imgur.com/qFur5.png" alt="enter image description here">

Of course, I can define it in main() function, but I wonder, what is a reason of this problem? Why cannot I declare pointer in header file and afterwards to include it in few source code files?

c++ header multiple variable-pointer inclusion

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我遇到了这样的问题：我有一个头文件：//first variant:  #ifndef LIBRARIES_H#define LIBRARIES_H const char a='4';     #endif // LIBRARIES_H我将其包含在几个.cpp文件中。一切都运行得很完美。但是突然，如果我声明一个指针：/...

问c++标头包含多个变量指针
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问c++标头包含多个变量指针EN