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What are the differences between pointer variable and reference variable in C++?
What's the meaning of * and & when applied to variable names?
Trying to understand meaning of "&" in this situation
void af(int& g) { g++; cout<<g; } If you call this function and pass variable name - it will act the same like normal void(int g). I know, when you write &g that means you are passing address of variable g. But what does it means in this sample?
It means you're passing the variable by reference.
In fact, in a declaration of a type, it means reference, just like:
int x = 42; int& y = x; declares a reference to x, called y.
The & means that the function accepts the address (or reference) to a variable, instead of the value of the variable.
For example, note the difference between this:
void af(int& g) { g++; cout<<g; } int main() { int g = 123; cout << g; af(g); cout << g; return 0; } And this (without the &):
void af(int g) { g++; cout<<g; } int main() { int g = 123; cout << g; af(g); cout << g; return 0; } 
void main() is fine. Nothing wrong with this. Of course I assume that this is incomplete code (he misses "#include" too), so there will be a namespace lulz { ... } around his void main.
int main to make Chris happy ;) And @LuchianGrigore: actually, yes, the function does accept the address. If you use printf("%x",&g); inside and outside of of af(), you will find the values are different (at least when I compile with gcc...). The compiler is actually making a copy of the memory address and passing that to the function; there IS another object on the stack, believe it or not (again, at least this is happening for me in gcc....)af(&g), right? Since &g is the address of g, and you say that af accepts the address of a variable?
af(v); af(v);?