What does "Memory allocated at compile time" really mean?

In programming languages like C and C++, people often refer to static and dynamic memory allocation. I understand the concept but the phrase "All memory was allocated (reserved) during compile time" always confuses me.

Compilation, as I understand it, converts high level C/C++ code to machine language and outputs an executable file. How is memory "allocated" in a compiled file ? Isn't memory always allocated in the RAM with all the virtual memory management stuff ?

Isn't memory allocation by definition a runtime concept ?

If I make a 1KB statically allocated variable in my C/C++ code, will that increase the size of the executable by the same amount ?

This is one of the pages where the phrase is used under the heading "Static allocation".

Back To Basics: Memory allocation, a walk down the history

Solution 1:

Memory allocated at compile-time means the compiler resolves at compile-time where certain things will be allocated inside the process memory map.

For example, consider a global array:

int array[100];

The compiler knows at compile-time the size of the array and the size of an int, so it knows the entire size of the array at compile-time. Also a global variable has static storage duration by default: it is allocated in the static memory area of the process memory space (.data/.bss section). Given that information, the compiler decides during compilation in what address of that static memory area the array will be.

Of course that memory addresses are virtual addresses. The program assumes that it has its own entire memory space (From 0x00000000 to 0xFFFFFFFF for example). That's why the compiler could do assumptions like "Okay, the array will be at address 0x00A33211". At runtime that addresses are translated to real/hardware addresses by the MMU and OS.

Value initialized static storage things are a bit different. For example:

int array[] = { 1 , 2 , 3 , 4 };

In our first example, the compiler only decided where the array will be allocated, storing that information in the executable.
In the case of value-initialized things, the compiler also injects the initial value of the array into the executable, and adds code which tells the program loader that after the array allocation at program start, the array should be filled with these values.

Here are two examples of the assembly generated by the compiler (GCC4.8.1 with x86 target):

C++ code:

int a[4];
int b[] = { 1 , 2 , 3 , 4 };

int main()
{}

Output assembly:

a:
    .zero   16
b:
    .long   1
    .long   2
    .long   3
    .long   4
main:
    pushq   %rbp
    movq    %rsp, %rbp
    movl    $0, %eax
    popq    %rbp
    ret

As you can see, the values are directly injected into the assembly. In the array a, the compiler generates a zero initialization of 16 bytes, because the Standard says that static stored things should be initialized to zero by default:

8.5.9 (Initializers) [Note]:
Every object of static storage duration is zero-initialized at program startup before any other initial- ization takes place. In some cases, additional initialization is done later.

I always suggest people to disassembly their code to see what the compiler really does with the C++ code. This applies from storage classes/duration (like this question) to advanced compiler optimizations. You could instruct your compiler to generate the assembly, but there are wonderful tools to do this on the Internet in a friendly manner. My favourite is GCC Explorer.

Solution 2:

Memory allocated at compile time simply means there will be no further allocation at run time -- no calls to malloc, new, or other dynamic allocation methods. You'll have a fixed amount of memory usage even if you don't need all of that memory all of the time.

Isn't memory allocation by definition a runtime concept?

The memory is not in use prior to run time, but immediately prior to execution starting its allocation is handled by the system.

If I make a 1KB statically allocated variable in my C/C++ code, will that increase the size of the executable by the same amount?

Simply declaring the static will not increase the size of your executable more than a few bytes. Declaring it with an initial value that is non-zero will (in order to hold that initial value). Rather, the linker simply adds this 1KB amount to the memory requirement that the system's loader creates for you immediately prior to execution.