Why does this little code consume extra bytes of memory?

In the following program, the structure object nd[0] node consumed 8 bytes of memory. Ideally, nd[0] should consume 5 bytes of memory in total. I mean, 4 bytes for Int type + 1 byte for the char type.

Code:

---

#include<iostream>
#include<stdio.h>
using namespace std;

struct Node
{
    int distance;
    char charName;
} nd[] = {5,'b'};


int main()
{
    cout << sizeof(nd[0]) << "  " << sizeof(nd[0].charName) << "    " << sizeof(nd[0].distance) << endl;
    return 0;
}

---

Output : 8 1 4

First of all, you cannot assume the order of the members of a struct, in memory, despite the order you specify.

If the char is first in memory, then the int will by default, align on the next int boundary, with 3 bytes of padding in between.

Your compiler will probably have options concerning the alignment of members of a struct. Unless you have limited memory, best to leave the alignment/padding up to the compiler.

In addition to all of this, although your question concerns both C & C++, this forum is for discussions of C, NOT C++

I suggest reading Eric Raymond's article on The Lost Art of Structure Packing. The section on "Structure alignment and padding" has a similiar example with explanation why there might be trailing padding by default.

Thanks Rstanley.

Thanks Laserlight.

> First of all, you cannot assume the order of the members of a struct, in memory, despite the order you specify.
Erm, yes you can.

Quote:

---

Within a structure object, the non-bit-field members and the units in which bit-fields
reside have addresses that increase in the order in which they are declared. A pointer to a
structure object, suitably converted, points to its initial member (or if that member is a
bit-field, then to the unit in which it resides), and vice versa. There may be unnamed
padding within a structure object, but not at its beginning.

---

Quote:

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Originally Posted by rstanley

First of all, you cannot assume the order of the members of a struct, in memory, despite the order you specify.

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You are mistaken:

Quote:

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Originally Posted by C11 (draft because I am on my phone) Clause 6.7.2.1 Paragraphs 13, 15

Within a structure object, the non-bit-field members and the units in which bit-fields reside have addresses that increase in the order in which they are declared. A pointer to a structure object, suitably converted, points to its initial member (or if that member is a bit-field, then to the unit in which it resides), and vice versa. There may be unnamed padding within a structure object, but not at its beginning.

There may be unnamed padding at the end of a structure or union.

---

I may have read some incorrect documentation many years ago.

I stand corrected.

Quote:

---

Originally Posted by Shafiul

In the following program, the structure object nd[0] node consumed 8 bytes of memory. Ideally, nd[0] should consume 5 bytes of memory in total. I mean, 4 bytes for Int type + 1 byte for the char type.

Code:

---

#include<iostream>
#include<stdio.h>
using namespace std;

struct Node
{
    int distance;
    char charName;
} nd[] = {5,'b'};


int main()
{
    cout << sizeof(nd[0]) << "  " << sizeof(nd[0].charName) << "    " << sizeof(nd[0].distance) << endl;
    return 0;
}

---

Output : 8 1 4

---

To explain what is going onn here abit more, you are guaranteed
that the first member will have the same address as the structure itself. You also guarantted that when you declare an array of structures, the size of the array is N * the size of the structure.

However most platforms have alignment requirements for multi-byte types like int. So it follows that to fulfil these demands, you need to insert three padding bytes, probably at the end of the struct through they could be after the fiirst member.