C + +: The misconception of macro substitution

Look at the following code, what is the result of the output?

#include <iostream>usingnamespacestd;#define NUM 0void fun(){    #undef NUM    #define NUM 100}int main(){    fun();    cout<<"NUM="<<NUM<<endl;//NUM=100;    return0;}

Yes, the answer is 100, and then look at the following code:

#include <iostream>usingnamespacestd;#define NUM 0void fun();int main(){    fun();    cout<<"NUM="<<NUM<<endl;//NUM=0;    return0;}void fun(){    #undef NUM    #define NUM 100}

The result of the output is 0, why? At this point I came to the preliminary conclusion that the macro replacement is not to scan the entire file and then replace all, in order to solve my question, I have the macro processing after the two function code map is intercepted.


It is obvious that the replacement result is not the same.

 #include <iostream>  using  namespace  STD ; void  fun ();    int  Main () {fun ();    Out; //main.cpp:7: Error: ' Out ' is not declared in this scope  //obviously made a mistake, because when the main function is not seen out.  return  0 ;} void  Fun () { #define out cout<< "Hello word" <<endl }//it can be concluded that macro substitution does not replace all macros at preprocessing time,   //macro defined after the main function is not expanded, the macro changes or macro definitions are not visible in main. 

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C + +: The misconception of macro substitution