##（双散列）在预处理器指令中做什么？ [英] What does ## (double hash) do in a preprocessor directive?

问题描述

  #define DEFINE_STAT（Stat）\ 
 struct FThreadSafeStaticStat< FStat _ ## Stat> StatPtr _ ## Stat; 
  

上面这行是从虚幻4，我知道我可以在虚幻的论坛，但我认为这是一个一般的C ++问题，值得被问到这里。

我理解第一行定义一个宏，但是我不熟悉预处理器shenanigans C ++，所以我迷失在那里。逻辑告诉我反斜杠意味着声明继续到下一行。

FThreadSafeStaticStat看起来有点像一个模板，但有＃在那里和一个语法I在C ++中从未见过

有人能告诉我这是什么意思吗？我知道你可能无法访问Unreal 4，但它只是我不懂的语法。

解决方案

<$

所以如果你使用

DEFINE_STAT（foo）

在代码中的任何位置，它将替换为

struct FThreadSafeStaticStat< FStat_foo>

这里是另一个例子： 我的博客帖子以进一步解释此问题。

  #include< stdio.h> 
 
 #define decode（s，t，u，m，p，e，d）m ## s ## u ## t 
 #define begin decode（a，n，i ，m，a，t，e）
 
 begin（）
 {
 printf（Stumped？\\\
）; 
} 
  

此程序将成功编译并执行，并生成以下输出： p>

  Stumped？ 
  

当预处理器处理此代码时，

< （a，n，i，m，a，t，e）替换为



• begin（） （）


• decode（a，n，i，m，a，t，e）（）替换为 m ## a ## i ## n（）


• m ## a ## i ## n（）替换为 main（）

有效地， begin（）替换为 main（）。

#define DEFINE_STAT(Stat) \
struct FThreadSafeStaticStat<FStat_##Stat> StatPtr_##Stat;

The above line is take from Unreal 4, and I know I could ask it over on the unreal forums, but I think this is a general C++ question that warrants being asked here.

I understand the first line defines a macro, however I am not well versed in preprocessor shenanigans in C++ and so I'm lost over there. Logic tells me the backslash means the declaration continues onto the next line.

FThreadSafeStaticStat looks a bit like a template, but there's #'s going on in there and a syntax I've never seen before in C++

Could someone tell me what this means? I understand that you may not have access to Unreal 4, but it's just the syntax I don't understand.

解决方案

## is the preprocessor operator for concatenation.

So if you use

DEFINE_STAT(foo)

anywhere in the code, it gets replaced with

struct FThreadSafeStaticStat<FStat_foo> StatPtr_foo;

before your code is compiled.

Here is another example from a blog post of mine to explain this further.

#include <stdio.h>

#define decode(s,t,u,m,p,e,d) m ## s ## u ## t
#define begin decode(a,n,i,m,a,t,e)

begin()
{
    printf("Stumped?\n");
}

This program would compile and execute successfully, and produce the following output:

Stumped?

When the preprocessor works on this code,

• begin() is replaced with decode(a,n,i,m,a,t,e)()
• decode(a,n,i,m,a,t,e)() is replaced with m ## a ## i ## n()
• m ## a ## i ## n() is replaced with main()

Thus effectively, begin() is replaced with main().

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