So lets say I have this code
int my_static_int = 4;
func(&my_static_int);
I passed the function a pointer to my_static_int, obviously. But what happens when the code is compiled? Avenue I’ve considered:
1) When you declare a non-pointer variable, C automatically creates its pointer and does something internally like typedefs my_static_int to be *(internal_reference)
Anyway, I hope that my question is descriptive enough
If you really want to know how the code looks under the covers, you have to get the compiler to generate the assembler code (
gcccan do this with the-Soption).When you truly grok C and pointers at their deepest level, you’ll realise that it’s just the address of the variable being passed in rather than the value of the variable. There’s no need for creating extra memory to hold a pointer since the pointer is moved directly from the code to the stack (the address will probably have been set either at link time or load time, not run time).
There’s also no need for internal type creation since the compiled code already knows the type and how to manipulate it.
Keeping in mind that this is implementation-specific, consider the following code:
which, when compiled with
gcc -Sto get the assembler, produces:The important bit is these sections:
Hence the address is passed, and used to get at the variable.