如何使C ++宏表现得像一个函数？ [英] How do I make a C++ macro behave like a function?

问题描述

让我们说，由于某种原因，你需要写一个宏： MACRO（X，Y）。 （假设有一个很好的理由不能使用内联函数。）您希望此宏模拟对没有返回值的函数的调用。

---

示例1：这应该能正常工作。

  if（x> y）
 MACRO（x，y）; 
 do_something（）; 
  

示例2：这不应导致编译器错误。

  if（x> y）
 MACRO（x，y）; 
 else 
 MACRO（y  -  x，x  -  y）; 
  

示例3：这不应该编译。

  do_something（）; 
 MACRO（x，y）
 do_something（）; 
  

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写宏的方式如下：

  #define MACRO（X，Y）\ 
 cout< < 第一个arg是：< （X） endl; \ 
 cout<< 第二个arg是：< （Y） endl; \ 
 cout<< Sum is：< （（X）+（Y））< endl; 
  

这是一个非常糟糕的解决方案，失败了所有三个例子，我不需要解释为什么

忽略宏的实际功能，

现在，我最常看到写的宏是用大括号括起来的，如下：

  #define MACRO（X，Y）\ 
 {\ 
 cout< 第一个arg是：< （X） endl; \ 
 cout<< 第二个arg是：< （Y） endl; \ 
 cout<< Sum is：< （（X）+（Y））< endl; \ 
} 
  

这解决了示例1，因为宏是在一个语句块中。但是示例2被打破，因为我们在调用宏之后放了一个分号。这使得编译器认为分号本身是一个语句，这意味着else语句不对应任何if语句！最后，示例3编译为OK，即使没有分号，因为代码块不需要分号。

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有没有办法写一个宏，以便它通过所有三个例子？

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注意：我正在提交自己的答案作为接受的方式共享一个提示，但如果任何人有一个更好的解决方案，感到自由地张贴在这里，它可能会获得比我的方法更多的投票。 ：）

解决方案

通常应避免宏;在任何时候都喜欢内联函数给他们。任何有价值的编译器都应该能够内联一个小函数，就像它是一个宏，内联函数将会尊重命名空间和其他范围，以及一次评估所有的参数。

如果它必须是一个宏，一个while循环（已经建议）将工作，或者你可以尝试逗号运算符：

  #define MACRO（X，Y）\ 
（\ 
（cout <1st arg is：<<（X）< ，\ 
（cout <<2nd arg is：<<（Y）<< endl），\ 
（cout< <（（X）+（Y））< -  endl），\ 
（void）0 \ 
）
  / pre> 
 
 
 （void）0 使语句评估为 void 类型，使用逗号而不是分号允许它在语句内部使用，而不是作为独立的。我仍然推荐一个内联函数的主要原因，其中最小的范围和事实， MACRO（a ++，b ++）会增加 a 和 b 两次。
 
Let's say that for some reason you need to write a macro: MACRO(X,Y).  (Let's assume there's a good reason you can't use an inline function.)  You want this macro to emulate a call to a function with no return value.



Example 1: This should work as expected.

if (x > y)
  MACRO(x, y);
do_something();


Example 2: This should not result in a compiler error.

if (x > y)
  MACRO(x, y);
else
  MACRO(y - x, x - y);


Example 3: This should not compile.

do_something();
MACRO(x, y)
do_something();


The naïve way to write the macro is like this:
#define MACRO(X,Y)                       \
cout << "1st arg is:" << (X) << endl;    \
cout << "2nd arg is:" << (Y) << endl;    \
cout << "Sum is:" << ((X)+(Y)) << endl;
This is a very bad solution which fails all three examples, and I shouldn't need to explain why.

Ignore what the macro actually does, that's not the point.



Now, the way I most often see macros written is to enclose them in curly braces, like this:
#define MACRO(X,Y)                         \
{                                          \
  cout << "1st arg is:" << (X) << endl;    \
  cout << "2nd arg is:" << (Y) << endl;    \
  cout << "Sum is:" << ((X)+(Y)) << endl;  \
}
This solves example 1, because the macro is in one statement block.  But example 2 is broken because we put a semicolon after the call to the macro.  This makes the compiler think the semicolon is a statement by itself, which means the else statement doesn't correspond to any if statement!  And lastly, example 3 compiles OK, even though there is no semicolon, because a code block doesn't need a semicolon.



Is there a way to write a macro so that it pass all three examples?



Note: I am submitting my own answer as part of the accepted way of sharing a tip, but if anyone has a better solution feel free to post it here, it may get more votes than my method. :)
解决方案
Macros should generally be avoided; prefer inline functions to them at all times. Any compiler worth its salt should be capable of inlining a small function as if it were a macro, and an inline function will respect namespaces and other scopes, as well as evaluating all the arguments once.

If it must be a macro, a while loop (already suggested) will work, or you can try the comma operator:
#define MACRO(X,Y) \
 ( \
  (cout << "1st arg is:" << (X) << endl), \
  (cout << "2nd arg is:" << (Y) << endl), \
  (cout << "3rd arg is:" << ((X) + (Y)) << endl), \
  (void)0 \
 )
The (void)0 causes the statement to evaluate to one of void type, and the use of commas rather than semicolons allows it to be used inside a statement, rather than only as a standalone. I would still recommend an inline function for a host of reasons, the least of which being scope and the fact that MACRO(a++, b++) will increment a and b twice.

                        
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