What does ## (double hash) do in a preprocessor directive?

## is the preprocessor operator for concatenation.

So if you use

DEFINE_STAT(foo)

anywhere in the code, it gets replaced with

struct FThreadSafeStaticStat<FStat_foo> StatPtr_foo;

before your code is compiled.

Here is another example from a blog post of mine to explain this further.

#include <stdio.h>#define decode(s,t,u,m,p,e,d) m ## s ## u ## t#define begin decode(a,n,i,m,a,t,e)int begin(){    printf("Stumped?\n");}

This program would compile and execute successfully, and produce the following output:

Stumped?

When the preprocessor is invoked on this code,

• begin is replaced with decode(a,n,i,m,a,t,e)
• decode(a,n,i,m,a,t,e) is replaced with m ## a ## i ## n
• m ## a ## i ## n is replaced with main

Thus effectively, begin() is replaced with main().

TLDR; ## is for concatenation and # is for stringification (from cppreference).

The ## concatenates successive identifiers and it is useful when you want to pass a function as a parameter. Here is an example where foo accepts a function argument as its 1st argument and the operators a and b as the 2nd and 3rd arguments:

#include <stdio.h>enum {my_sum=1, my_minus=2};#define foo(which, a, b) which##x(a, b)#define my_sumx(a, b) (a+b)#define my_minusx(a, b) (a-b)int main(int argc, char **argv) {    int a = 2;    int b = 3;    printf("%d+%d=%d\n", a, b,  foo(my_sum, a, b));  // 2+3=5    printf("%d-%d=%d\n", a, b, foo(my_minus, a, b)); // 2-3=-1    return 0;}

The # concatenates the parameter and encloses the output in quotes. The example is:

#include <stdio.h> #define bar(...) puts(#__VA_ARGS__)int main(int argc, char **argv) {    bar(1, "x", int); // 1, "x", int    return 0;}