Why is a datatype necessary for an initialization that follows a declaration?

Question

Consider the simple C program:

int a;      // declaration
int a = 11; // initialization

int main(int argc, char* argv[]) {
    int b;  // declaration
    b = 10; // assignment

If the initialization of a were written without the data type, such as a = 11, the compiler raises a warning. Why does the initialization of a require a data type, when the declaration of a already specifies its data type?

Answer 1

int a at file scope is a "tentative" definition (because it lacks the initialising part). This means that a can be defined again with a value at a later point.

A tentative definition may or may not act as a definition, depending on whether there is an actual external definition before or after it in the translation unit:

int a = 5; // defines a in the current translation unit with external linkage and a value of 5
int a; // tentative definition with no effect (a is already defined)

The other way around usually has more practical merit:

int a;
...
int a = 5;

The tentative definition could precede the actual definition, if, for example, the constant used to initialise it is not available at the first point.

Edit:

You seem to be confused that you are not able perform an assignment at file scope. A C program is allowed to have actual operations only within functions. At file scope, one may only define or declare variables, types and functions.

Answer 2

I think this has something to do with the fact that you can't write instructions in the global scope. What it means is :

int a = 11;

Defines a variable. This tells the compiler to assign a static address to the variable, because it is global. The default (assignment) value is just an added bonus. Whereas :

a = 11;

Is an instruction, which is illegal.