C ++宏有条件编译代码？ [英] C++ Macro to conditionally compile code?

问题描述

我想根据宏来有条件地编译代码。基本上我有一个宏，看起来像（从真实版本简化）：

  #if DEBUG 
 #define START_BLOCK （x）if（DebugVar（#x）\ 
 {char debugBuf [8192]; 
 #define END_BLOCK（）printf（％s\\\
，debugBuf）;} 
 #else 
 #define START_BLOCK（x）（void）0; 
 #define END_BLOCK（）（void）0; 
 #endif 
 问题是如果 DEBUG 被定义，你可以这样做：
  START_BLOCK（test）
 char str [] =Test is defined; 
 strcpy（debugBuf，str）; 
 END_BLOCK（）
 
 START_BLOCK（foo）
 char str [] =Foo is defined; 
 strcpy（debugBuf，str）; 
 END_BLOCK b $ b  
一切都正常，因为每个块都在自己的范围内，但如果DEBUG没有定义，你会得到一个重新定义 str 在第二个块。（你也会得到 debugBuf 没有定义，这只是简化示例的一个副作用。）
 
 
 
我想做的是让#else像这样：
  #else 
 #define START_BLOCK（x）#if 0 
 #define END_BLOCK（）#endif 
 #endif 
  
或者在开始/结束块之间没有任何东西的其他方法被编译。我试过上面的，我也尝试了一些类似的行：
  #else 
 #define NULLMACRO 。）（void）0 
 #define START_BLOCK（x）NULLMACRO（
 #define END_BLOCK（））
 #endif 
  
没有运气。
 
 
 
有办法吗？我想到的只是我可能滥用优化编译器和使用：
  #else 
＃定义START_BLOCK（x）if（0）{
 #define END_BLOCK（）} 
 #endif 
  
并相信它会完全编译块。 
解决方案
那么你想要有自己的范围的条件块吗？
 
 
 
这是一个相当可读的解决方案，依靠编译器来优化它：
  #define DEBUG 1 
 
 if（DEBUG）{
 // ... 
} 
  
这里是一个只有预处理器的：
  #define DEBUG 1 
 
 #ifdef DEBUG 
 #define IFDEBUG（x）{x} 
 #else 
 #define IFDEBUG（x）
 #endif 
 
 IFDEBUG b $ b // ... 
）
  
或手动：
  #define DEBUG 1 
 
 #ifdef DEBUG 
 {
 // ... 
} 
 #endif 
  
 
I want to compile code conditionally based on a macro. Basically I have a macro that looks like (Simplified from the real version):
#if DEBUG
    #define START_BLOCK( x ) if(DebugVar(#x) \
        { char debugBuf[8192];
    #define END_BLOCK( ) printf("%s\n", debugBuf); }
#else
    #define START_BLOCK( x ) (void)0;
    #define END_BLOCK( ) (void)0;
#endif
The issue is that if DEBUG is defined you could do things like:
START_BLOCK( test )
     char str[] = "Test is defined";
     strcpy(debugBuf, str);
END_BLOCK( )

START_BLOCK( foo )
    char str[] = "Foo is defined";
    strcpy(debugBuf, str);
END_BLOCK( )
And everything works fine because each block is within it's own scope. However if DEBUG isn't defined, then you'd get a redefinition of str in the second block. (Well you'd also get debugBuf not defined but that's just a side effect of the simplified example.)

What I'd like to do is to have the #else be something like:
#else
    #define START_BLOCK( x ) #if 0
    #define END_BLOCK( ) #endif
#endif
Or some other method of not having anything between the start / end blocks be compiled. I tried the above, I also tried something along the lines of:
#else
    #define NULLMACRO( ... ) (void)0
    #define START_BLOCK( x ) NULLMACRO(
    #define END_BLOCK( ) )
#endif
without any luck.

Is there a way for this to work? One thought that just occurred to me is that I could maybe abuse the optimizing compiler and use:
#else
    #define START_BLOCK( x ) if(0){
    #define END_BLOCK( ) }
#endif
And trust that it will just compile out the block completely. Are there any other solutions?
解决方案
So you want conditional blocks with their own scope?

Here's a quite readable solution that relies on the compiler to optimize it away:
#define DEBUG 1

if (DEBUG) {
    // ...
}
And here is one that is preprocessor-only:
#define DEBUG 1

#ifdef DEBUG
    #define IFDEBUG(x) {x}
#else
    #define IFDEBUG(x)
#endif

IFDEBUG(
    // ...
)
Or manually:
#define DEBUG 1

#ifdef DEBUG
{
    // ...
}
#endif


                        
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