The meaning of # ## in C++/C macro definition (define)

Original link: http://blog.csdn.net/qq_15457239/article/details/56842450

#define f(a,b) a##b
#define d(a) #a
#define s(a) d(a)

void main( void )
{
puts(d(f(a,b)));
puts(s(f(a,b)));
}

输出结果:
f(a,b)
ab

Analysis: ##Connecting two symbols
#a means treating a as a symbol, that is, treating the following # as a string

The # and ## operators are used with the #define macro. Use # to return the first argument after # as a quoted string. For example, the command
    #define to_string( s ) # s
will cause the compiler to The following command
    cout < < to_string( Hello World! ) < < endl;
is understood as
    cout < < "Hello World!" < < endl;
Use ## to concatenate the content before and after ##. For example, the command
    #define concatenate( x, y ) x ## y
...
int xy = 10;
... will
cause the compiler to interpret
    cout < < concatenate( x, y ) < < endl;
as
    cout < < xy < < endl;
'10' is displayed at the output.

puts(d(f(a,b))); ----> 因为d宏中的参数是另外一个宏，且带##，所以作为参数的宏不展开，相当于
puts(#f(a,b));----->puts("f(a,b)");
puts(s(f(a,b))); ----> 因为s宏中的参数是另外一个宏，但不带##，所以作为参数的宏先展开，相当于
puts(s(ab));----->puts(d(ab));---->puts(#ab);---->puts("ab");

#define f(a,b) a##b
#define d(a) #a －－》以"#"开头的，直接替换，不展开：immediately replaced by the unexpanded actual argument
#define s(a) d(a) －－》非以"#"开头的，先展开，再替换，也就是一般的情况

所以就两种情况：
1，不以"#"开头的，先展开参数a，然后是替换代码：puts(s(f(a,b)));-->puts(s(ab))-->puts(d(ab))-->puts("ab")
2，以"#"开头的，直接替换，不展开：puts(d(f(a,b)))-->puts("f(a,b)")

原文链接：http://blog.csdn.net/qq_15457239/article/details/56842450

#define f(a,b) a##b
#define d(a) #a
#define s(a) d(a)

void main( void )
{
puts(d(f(a,b)));
puts(s(f(a,b)));
}

输出结果:
f(a,b)
ab

分析： ##把两个符号连起来
#a指把a当成符号，就是把#后面的看成字符串

# 和 ## 操作符是和#define宏使用的. 使用# 使在#后的首个参数返回为一个带引号的字符串. 例如, 命令
    #define to_string( s ) # s
将会使编译器把以下命令
    cout < < to_string( Hello World! ) < < endl;
理解为
    cout < < "Hello World!" < < endl;
使用##连结##前后的内容. 例如, 命令
    #define concatenate( x, y ) x ## y
...
int xy = 10;
...
将会使编译器把
    cout < < concatenate( x, y ) < < endl;
解释为
    cout < < xy < < endl;
理所当然,将会在标准输出处显示'10'.

The meaning of # ## in C++/C macro definition (define)

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