cppreference Replacing text macros

# and ## operators

NOTE:

1、原文的内容， 非常冗长，不易阅读理解

# stringify and ## concatenate

关于此，参见:

gsl-lite/include/gsl/gsl-lite.hpp

#define  gsl_lite_MAJOR  0
#define  gsl_lite_MINOR  38
#define  gsl_lite_PATCH  0

#define  gsl_lite_VERSION  gsl_STRINGIFY(gsl_lite_MAJOR) "." gsl_STRINGIFY(gsl_lite_MINOR) "." gsl_STRINGIFY(gsl_lite_PATCH)

#define gsl_STRINGIFY(  x )  gsl_STRINGIFY_( x )
#define gsl_STRINGIFY_( x )  #x
#define gsl_CONCAT_(  a, b )  gsl_CONCAT2_( a, b )
#define gsl_CONCAT2_( a, b )  a##b

stackoverflow # and ## in macros

NOTE:

1、非常好的例子

#include <stdio.h>
#define f(a,b) a##b
#define g(a)   #a
#define h(a) g(a)

int main()
{
    printf("%s\n", h(f(1,2)));
    printf("%s\n", g(f(1,2)));
    return 0;
}

NOTE:

下面是我的测试:

#define f(a,b) a##b
#define g(a)   #a
#define h(a) g(a)

int main()
{
  printf("%s\n", h(f(1,2)));
  printf("%s\n", g(f(1,2)));
  return 0;
}

gcc -E test.cpp

# 1 "test.cpp"
# 1 "<built-in>"
# 1 "<命令行>"
# 1 "/usr/include/stdc-predef.h" 1 3 4
# 1 "<命令行>" 2
# 1 "test.cpp"




int main()
{
 printf("%s\n", "12");
 printf("%s\n", "f(1,2)");
 return 0;
}

A

NOTE:

1、这个解释非常好

An occurrence of a parameter in a function-like macro, unless it is the operand of # or ##, is expanded before substituting it and rescanning the whole for further expansion. Because g's parameter is the operand of #, the argument is not expanded but instead immediately stringified ("f(1,2)"). Because h's parameter is not the operand of # nor ##, the argument is first expanded (12), then substituted (g(12)), then rescanning and further expansion occurs ("12").

A

Because that is how the preprocessor works.

A single '#' will create a string from the given argument, regardless of what that argument contains, while the double '##' will create a new token by concatenating the arguments.

Try looking at the preprocessed output (for instance with gcc -E) if you want to understand better how the macros are evaluated.

Example

1、SergiusTheBest/ScopeExit

2、More C++ Idioms/Member Detector