c++ variable does not name a type in macro

Question

Have this code:

#include <iostream>

int a=0;

#define F(f) \
  int t##f(int, int);\
  a ++;\
  int t##f(int i, int j)  

F(nn) {
    return i*j;
}

int main() {
 int b = tnn(3, 8);
 std::cout << a << b;
}

Got error when compiling:

7:3: error: 'a' does not name a type
10:1: note: in expansion of macro 'F'

Why isn't a visible in macro at the position it expands?

Answer 1

Look at the expansion of the macro:

F(nn) becomes

int tnn(int, int);
a++;
int tnn(int i, int j) {
  return i * j;
}

The variable 'a' is being incremented outside a function which is a syntax error.

Like the other answer said, you cannot execute statements wherever you please; statements must be inside a function in order to be valid.

There are a few things that can go in the global scope :

1. Namespace declaration and definitions
2. Global variable declarations
3. Function prototypes and definitions
4. Template and class declarations and definitions
5. Preprocessor directives

Things that must be in function scope:

1. Control statements such as if and for
2. Labels
3. Function calls

Finally, the above lists are not all inclusive.

Answer 2

Your macro ( in the nn case) expands to:

int a=0;

int tnn(int, int); a ++; int tnn(int i, int j)  {
    return i*j;
}

int main() {
 int b = tnn(3, 8);
 std::cout << a << b;
}

There is no global scope in C++. That is only in scripting languages. Execution order is an initialization library-- something like crt0.s which constructs your run time envioronment. Then initialize global variables ( this part can get very complicated ) then run main.

You statement fails simply because you cannot put arbitrariy executable code where the macro is expanded.

PS: Bjarne says don't use macros. In fact he create const, inline and to some degree templates so that you can avoid macros. Macros are evil!!!!