What do # and ## operators do in C? [duplicate]

Answer 1

operator ## concatenates two arguments leaving no blank spaces between them..

 #define printe(a,b) a ## b
   printe(c,out) << "testing";

output is : testing

and single # is used for parameter replacement withe the string parameter like

#define  st(x)  #x
 cout<<st(tesing); //  equivalent to  cout<<"testing";

and # is also a preprocessor directive..

Answer 2

The '#' is really not an operator, they are preprocessor directives, and the '##' is used only for function macro definitions.

---

There are many preprocessor directives in C:

For Macro Definitions there are:

#define
#undef

For Conditional Inclusions, there are:

#ifdef
#ifndef
#if
#endif
#else
#elif

For Line Control, there is:

#line

For Error, there is:

#error

For Source file inclusion, there is:

#include

For Pragma directive, there is:

#pragma

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For more information, read this http://www.cplusplus.com/doc/tutorial/preprocessor/

Answer 3

A code statement beginning with # indicates what follows is a preprocessor directive and should be expanded by the pre-processor.

## is called token Pasting or Token concatenation macro.

Answer 4

From the wikipedia page describing the C preprocessor:

The ## operator concatenates two tokens into one token, as in this example:

#define DECLARE_STRUCT_TYPE(name) typedef struct name##_s name##_t
DECLARE_STRUCT_TYPE(g_object); // Outputs typedef struct g_object_s g_object_t;

The # operator signals other directives to the C preprocessor, for example: #include, #define, #undef, #error, #if, #ifdef, #ifndef, #else, #elif, #endif