Size of structure with a char, a double, an int and a t [duplicate]

Question

When I run only the code fragment

int *t;
std::cout << sizeof(char)   << std::endl;
std::cout << sizeof(double) << std::endl;
std::cout << sizeof(int)    << std::endl;
std::cout << sizeof(t)      << std::endl;

it gives me a result like this:

1
8
4
4

Total: 17.

But when I test sizeof struct which contains these data types it gives me 24, and I am confused. What are the additional 7 bytes?

This is the code

#include <iostream>
#include <stdio.h>
struct struct_type{
    int i;
    char ch;
    int *p;
    double d;
} s;

int main(){
    int *t;
    //std::cout << sizeof(char)   <<std::endl;
    //std::cout << sizeof(double) <<std::endl;
    //std::cout << sizeof(int)    <<std::endl;
    //std::cout << sizeof(t)      <<std::endl;

    printf("s_type is %d byes long",sizeof(struct struct_type));

    return 0;
}

:EDIT

I have updated my code like this

#include <iostream>
#include <stdio.h>
struct struct_type{
    double d_attribute;
    int i__attribute__(int(packed));
    int * p__attribute_(int(packed));;
    char  ch;
} s;

int main(){
    int *t;
    //std::cout<<sizeof(char)<<std::endl;
    //std::cout<<sizeof(double)<<std::endl;
    //std::cout<<sizeof(int)<<std::endl;
    //std::cout<<sizeof(t)<<std::endl;

    printf("s_type is %d bytes long",sizeof(s));

    return 0;
}

and now it shows me 16 bytes. Is it good, or have I lost some important bytes?

Answer 1

... it gives me 24, and I am confused. What are the additional 7 bytes?

These are padding bytes inserted by the compiler. Data structure padding is implementation dependent.

From Wikipedia, Data structure alignment:

Data alignment means putting the data at a memory offset equal to some multiple of the word size, which increases the system's performance due to the way the CPU handles memory. To align the data, it may be necessary to insert some meaningless bytes between the end of the last data structure and the start of the next, which is data structure padding.

Answer 2

There is some unused bytes between some members to keep the alignments correct. For example, a pointer by default reside on 4-byte boundaries for efficiency, i.e. its address must be a multiple of 4. If the struct contains only a char and a pointer

struct {
  char a;
  void* b;
};

then b cannot use the adderss #1 — it must be placed at #4.

  0   1   2   3   4   5   6   7
+---+- - - - - -+---------------+
| a | (unused)  | b             |
+---+- - - - - -+---------------+

In your case, the extra 7 bytes comes from 3 bytes due to alignment of int*, and 4 bytes due to alignment of double.

  0   1   2   3   4   5   6   7   8   9   a   b   c   d   e   f
+---------------+---+- - - - - -+---------------+- - - - - - - -+
| i             |ch |           | p             |               |
+---------------+---+- - - - - -+---------------+- - - - - - - -+
 10  11  12  13  14  15  16  17
+-------------------------------+
| d                             |
+-------------------------------+