ç$ P $＃pprocessor和##运营商 [英] C preprocessor # and ## operators

问题描述

借助 C99标准文档已经在部分下面的例子涉及到## preprocessing运营商：

The C99 standard document has the following example in the section related to the ## preprocessing operator:

在下面的代码片段：

#define hash_hash # ## #
#define mkstr(a) # a
#define in_between(a) mkstr(a)
#define join(c, d) in_between(c hash_hash d)

char p[] = join(x, y); // equivalent to
                       // char p[] = "x ## y";

  
  

扩大生产，在不同的
  阶段：

The expansion produces, at various stages:

join(x, y)
in_between(x hash_hash y)
in_between(x ## y)
mkstr(x ## y)
"x ## y"

在换句话说，扩大hash_hash
  产生一个新的标记，由
  两个相邻的尖锐的迹象，但这种新的
  令牌不是##运营商。

In other words, expanding hash_hash produces a new token, consisting of two adjacent sharp signs, but this new token is not the ## operator.

我不明白为什么hash_hash的替代产生##，而不是##或＃，＃。什么样的作用是前后双哈希播放？

I don't understand why the substitution of hash_hash produces ## and not "##" or "#""#". What role are the single hashes before and after the double hash playing?

任何响应大大AP​​ preciated。

Any responses greatly appreciated.

推荐答案

的 ## 在＃##＃就像在这个前pression转义序列。它加到最左边和最右边的＃来最终产生令牌 ## 。简单地定义宏为 ## 将导致一个错误，因为连接运算符需要两个操作数。

The ## in # ## # acts like an escape sequence in this expression. It concatenates the leftmost and the rightmost # to finally produce the token ##. Simply defining the macro as ## would cause an error since the concatenation operator expects two operands.

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