Understand the compile time operator: sizeof

During I read the source code of Redis, I found the following code:

dict *d = zmalloc(sizeof(*d));

After searching a different definition of ‘d’, I realized that the d is the same object defined in the same line. In order to understand what happens, I did some testing:

#include <stdio.h>
#include <malloc.h>

typedef struct {
    int x;
    int y;
} dict;

int main() {
    dict *d = malloc(sizeof(*d));
    printf("sizeof d: %d, *d: %d, type: %dn", 
           sizeof(d), sizeof(*d), sizeof(dict));

    free(d);
    return 0;
}

It works as expected, I don’t understand how does it happen at run time. In order to find the magic, I deassembled the executable file by objdump, and realized that: the sizeof is a compile time operator (Which maybe widely known, but I didn’t know that), and all the sizeof has been calculated during compilation. Here’s the part of objdump result of the executable file:

$ objdump -xsd sizeof_test
00000000004005ac <main>:
  //save the previous rbp on the stack
  4005ac: 55              push   %rbp     
  //rbp == stack pointer at the begining  
  4005ad: 48 89 e5        mov    %rsp,%rbp
  //rbp takes 8 bytes, and the only local variable which is a pointer 
  //also takes 8 bytes, so make room for 16 bytes
  4005b0: 48 83 ec 10     sub    $0x10,%rsp

  //want to malloc 8 bytes
  4005b4: bf 08 00 00 00  mov    $0x8,%edi
  //call malloc function
  4005b9: e8 d2 fe ff ff  callq  400490 <malloc@plt>
  //assigned the returned value from malloc to the first local variable 
  //which is d
  4005be: 48 89 45 f8     mov    %rax,-0x8(%rbp)