Why does this C code compile?

Question

#include <stdio.h> int main() {     int c = c;     printf("c is %i\n", c);     return 0; } 

I'm defining an integer variable called c, and I'm assigning its value to itself. But how can this even compile? c hasn't been initialized, so how can its value be assigned to itself? When I run the program, I get c is 0.

I am assuming that the compiler is generating assembly code that is assigning space for the the c variable (when the compiler encounters the int c statement). Then it takes whatever junk value is in that un-initialized space and assigns it back to c. Is this what's happening?

Answer 1

I remember quoting this in a previous answer but I can't find it at the moment.

C++03 §3.3.1/1:

The point of declaration for a name is immediately after its complete declarator (clause 8) and before its initializer (if any), ...

Therefore the variable c is usable even before the initializer part.

Edit: Sorry, you asked about C specifically; though I'm sure there is an equivalent line in there. James McNellis found it:

C99 §6.2.1/7: Any identifier that is not a structure, union, or enumeration tag "has scope that begins just after the completion of its declarator." The declarator is followed by the initializer.

Answer 2

Your guess is exactly right. int c pushes space onto the stack for the variable, which is then read from and re-written to for the c = c part (although the compiler may optimize that out). Your compiler is pushing the value on as 0, but this isn't guaranteed to always be the case.