What's the meaning of “##” in a C++ macro? [duplicate]

Question

This question already has an answer here:

• What are the applications of the ## preprocessor operator and gotchas to consider? 13 answers
• The ## operator in C 7 answers

What's the meaning of "##" in the following?

#define CC_SYNTHESIZE(varType, varName, funName)\
protected: varType varName;\
public: inline varType get##funName(void) const { return varName; }\
public: inline void set##funName(varType var){ varName = var; }

Answer 1

The operator ## concatenates two arguments leaving no blank spaces between them: e.g.

#define glue(a,b) a ## b
glue(c,out) << "test";

This would also be translated into:

cout << "test";

Answer 2

It concatenates tokens without leaving blanks between them. Basically, if you didn't have the ## there

public: inline varType getfunName(void) const { return varName; }\

the precompiler would not replace funName with the parameter value. With ##, get and funName are separate tokens, which means the precompiler can replace funName and then concatenate the results.

Answer 3

This is called token pasting or token concatenation.

The ## (double number sign) operator concatenates two tokens in a macro invocation (text and/or arguments) given in a macro definition.

Take a look here at the official GNU GCC compiler documentation for more information.