C ++定义表达式求值 [英] C++ define expression evaluation

问题描述

假设我们有这个表达式：

  #define cube（x）x * x * x 
  

然后我们称之为：

  int n = 3，v; 
 v = cube（n + 1）; // v = 10 
 v = cube（（n + 1））; // v = 64 
 v = cube（n）; // v = 27 
  

所以问题是：为什么第一个操作不会使 v = 64 ？

解决方案

宏不进行评估

在编译文件之前，还有一个名为 C预处理器以文字方式替换宏调用 / 并准备

  #define cube（x）x * x * x当你这样做
  

此

  v = cube（n + 1）; 
 

替换为此（ expaned是正确的字词）

  v = n + 1 * n + 1 * n + 1; 
 //简化为
 v = n + n + n + 1; 
 //和
 v = 3 * n + 1; 
 

n = 3 code> 10 正确的观察结果。

注意，当你添加括号

  v = cube（（n + 1））; 
  

然后，扩展名为

  v =（n + 1）*（n + 1）*（n + 1） 
  

这是你期望的 cube（）

  #define cube（x）（（x）*如果使用 gcc ，请尝试   / p> 
 
 
  gcc -E source.c 
  
并检查结果以验证宏是如何扩展的。
 
Suppose we have this expression:
#define cube(x) x * x * x
And then we call it:
int n = 3, v;
v = cube(n + 1);   // v = 10
v = cube((n + 1)); // v = 64
v = cube(n);       // v = 27
So the question is: why first operation do not make v = 64?
解决方案
Macros are not evaluated (in the sense of the common interpretation of evaluation), they are expanded at compile time.

Before the file is compiled, there is another program called the C Preprocessor that replaces the macro invocation literally/textually and prepares the file for actual compilation, so for your macro 
#define cube(x) x * x * x when you do this
This
v = cube(n + 1);
is replaced with this (expaned is the correct term)
v = n + 1 * n + 1 * n + 1;
// Simplifies to
v = n + n + n + 1;
// and again
v = 3 * n + 1;
which for n = 3 gives you 10 exactly the observed result.

Note, that when you add parentheses
v = cube((n + 1));
then, the expansion is
v = (n + 1) * (n + 1) * (n + 1);
which is what you would expect cube() to do, so prevent this you should redefine your macro like this
#define cube(x) ((x) * (x) * (x))
If you are using gcc try
gcc -E source.c
and check the result to verify how the macro was expanded.

                        
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