I know when a program is run, the main() function is executed first. But when does the initialization of global variables declared outside the main() happens? I mean if I declare a variable like this:
unsigned long current_time = millis();
void main() {
while () {
//some code using the current_time global variable
}
}
Here, the exact time when the global variable initializes is important. Please tell what happens in this context.
Since you didn't define the language you're talking about, I assumed it to be C++.
In computer programming, a global variable is a variable that is accessible in every scope (unless shadowed). Interaction mechanisms with global variables are called global environment (see also global state) mechanisms. The global environment paradigm is contrasted with the local environment paradigm, where all variables are local with no shared memory (and therefore all interactions can be reconducted to message passing). Wikipedia.
In principle, a variable defined outside any function (that is, global, namespace, and class static variables) is initialized before main() is invoked. Such nonlocal variables in a translation unit are initialized in their declaration order (§10.4.9). If such a variable has no explicit initializer, it is by default initialized to the default for its type (§10.4.2). The default initializer value for built-in types and enumerations is 0. [...] There is no guaranteed order of initialization of global variables in different translation units. Consequently, it is unwise to create order dependencies between initializers of global variables in different compilation units. In addition, it is not possible to catch an exception thrown by the initializer of a global variable (§14.7). It is generally best to minimize the use of global variables and in particular to limit the use of global variables requiring complicated initialization. See.
(Quick answer: The C standard doesn't support this kind of initialization; you'll have to consult your compiler's documentation.)
Now that we know the language is C, we can see what the standard has to say about it.
C99 6.7.8 paragraph 4:
All the expressions in an initializer for an object that has static storage duration shall be constant expressions or string literals.
And the new 2011 standard (at least the draft I has) says:
All the expressions in an initializer for an object that has static storage duration shall be constant expressions or string literals.
So initializing a static object (e.g., a global such as your current_time) with a function call is a constraint violation. A compiler can reject it, or it can accept it with a warning and do whatever it likes if it provides an language extension.
The C standard doesn't say when the initialization occurs, because it doesn't permit that kind of initialization. Basically none of your code can execute before the main() function starts executing.
Apparently your compiler permits this as an extension (assuming you've actually compiled this code). You'll have to consult your compiler's documentation to find out what the semantics are.
(Normally main is declared as int main(void) or int main(int argc, char *argv[]) or equivalent, or in some implementation-defined manner. In many cases void main() indicates a programmer who's learned C from a poorly written book, of which there are far too many. But this applies only to hosted implementations. Freestanding implementations, typically for embedded systems, can define the program's entry point any way they like. Since you're targeting the Arduino, you're probably using a freestanding implementation, and you should declare main() however the compiler's documentation tells you to.)
