What does a function name indicate and more - c++

Question

Consider this simple program:
#include <iostream>
void length(void){
std::cout<<"My length is void"<<std::endl;
}
void width(void){
std::cout<<"My width is void"<<std::endl;
}
int main(void){
std::cout<<"The program length is: "<<length<<std::endl;
std::cout<<"The program width is: "<<width<<std::endl;
}
The output of this program is:
The program length is: 1
The program width is: 1
What are the numbers that the program prints out, and basically not having much knowledge in C++ particularly, this looks more pythonic syntax to me,and looks as if it should print the function address. I say this because, this problem arose initially while practicing some very basic programs for openGL,
The callback was registered in GLUT with something like:
void line(void){
//some code for line program ...
}
int main(int argc, char** argv){
//some more code;
glutDisplayFunc(line);
glutMainLoop();
}
It's almost seeming as if we're passing the address of the function, but, clearly from the above program this is not an address, and the syntax for pointer to function is a bit different, if so, how is this function being registered for callback? And what is it that we're passing to glutDisplayFunc?
and, since I **wanted to register a function that had passed arguments, I search a C++ analogy for python lambda functions, and found similar lambda function, but it didn't work out: **
#include <iostream>
void line(int a, int b){
//some code that does some plotting in the display window
}
int main(int argc, char** argv){
auto callback = [](void)->void{
line(a,b);
};
glutDisplayFunc(callback);
glutMainLoop();
}
This totally fails,the error that is shown is:
no suitable conversion function from "lambda []void ()->void" to "void (*)()" exists
This was using my python analogy, but what's the solution to this problem? The major parts of confusion are highlighted in bold.

Answer 1

It's almost seeming as if we're passing the address of the function, but, clearly from the above program this is not an address
You've got things backwards. It's not glutDisplayFunc that's doing the weird magical thing; it's std::cout.
Yes, a function name will convert to a function pointer. The combination of operator<< overloads for iostreams and C++ overload resolution rules ultimately leads to << treating functions as if they were boolean values (yes, really). And thus, your << length is equivalent to doing << true.
If you want the address, you can cast the function pointer to a void* before outputting it: << reinterpret_cast<void*>(&length) or the more terse << (void*)&length (though technically this is only conditionally supported by C++, pretty much every real compiler will allow this).
Lastly, in C++, lambdas are not functions; they're objects. You cannot pass an object to something that expects a function pointer. You can convert a capture-less lambda into a function pointer, but you can't do that for one which captures values.