Do functions occupy memory space?

Question

void demo()
{
    printf("demo");
}

int main()
{
    printf("%p",(void*)demo);
    return 0;
}

The above code prints the address of function demo.
So if we can print the address of a function, that means that this function is present in the memory and is occupying some space in it.
So how much space it is occupying in the memory?

Answer 1

You can see for yourself using objdump -r -d:

0000000000000000 <demo>:
   0:   55                      push   %rbp
   1:   48 89 e5                mov    %rsp,%rbp
   4:   bf 00 00 00 00          mov    $0x0,%edi
                        5: R_X86_64_32  .rodata
   9:   b8 00 00 00 00          mov    $0x0,%eax
   e:   e8 00 00 00 00          callq  13 <demo+0x13>
                        f: R_X86_64_PC32        printf-0x4
  13:   5d                      pop    %rbp
  14:   c3                      retq   

0000000000000015 <main>:

EDIT

I took your code and compiled (but not linked!) it. Using objdump you can see the actual way the compiler lays out the code to be run. At the end of the day there is no such thing as a function: for the CPU it's just a jump to some location (that in this listing happens to be labeled). So the size of the "function" is the size of the code that comprises it.

---

There seems to be some confusion that this is somehow not "real code". Here is what GDB says:

Dump of assembler code for function demo:
   0x000000000040052d <+0>:     push   %rbp
   0x000000000040052e <+1>:     mov    %rsp,%rbp
   0x0000000000400531 <+4>:     mov    $0x400614,%edi
   0x0000000000400536 <+9>:     mov    $0x0,%eax
   0x000000000040053b <+14>:    callq  0x400410 <printf@plt>
   0x0000000000400540 <+19>:    pop    %rbp
   0x0000000000400541 <+20>:    retq   

This is exactly the same code, with exactly the same size, patched by the linker to use real addresses. gdb prints offsets in decimal while objdump uses the more favourable hex. As you can see, in both cases the size is 21 bytes.