Getting the size in bytes or in chars of a member of a struct or union in C/C++?

Question

Let's say that I want to get the size in bytes or in chars for the name field from:

struct record
{
    int id;
    TCHAR name [50];
};

sizeof(record.name) does not work.

Answer 1

In C++:

#include <iostream>
using namespace std;;

struct record
{
    int id;
    char name [50];
};

int main() {
    cout << sizeof( record::name) << endl;
}

Edit: A couple of people have pointed out that this is C++0x code, so I guess I must retract my unkind comment regarding VC++. This is not a programming construct I have ever used in my own C++ code, but I have to wonder why sizeof would not work this way in C++03? You hand it a name and it gives you the size. I'd have thought it would take some effort for it not to work. But such is the wonder of the C++ Standard :-)

Answer 2

The solution for this is not so pretty as you may think:

size_in_byte = sizeof(((struct record *) 0)->name)

size_in_chars = _countof(((struct record *) 0)->name)

If you want to use the second one on other platforms than Windows try:

#define _countof(array) (sizeof(array)/sizeof(array[0]))

Answer 3

In C++, the syntax is sizeof(record::name).

Edit: as Adrian pointed out, this is valid C++0x, but not C++98. In that case, sizeof(record().name) is valid as long as record is default-constructible.

C is trickier, since there is no scope-resolution operator. The best I can think of is

typedef TCHAR record_name[50];
struct record
{
    int id;
    record_name record;
};

sizeof(record_name);

Edit: or, as Sorin suggested sizeof(((struct record*)0)->name)

Answer 4

If you create an instance first, it will work.

record r;
sizeof(r.name);

Answer 5

record is the name of a type, but record.name is not. You somehow have to access name through an instance of the struct. Sorin's answer is the usual C solution:

sizeof ((struct record*)0)->name;

This creates a pseudo-pointer to an instance (or pointer to a pseudo-instance) of struct record, then access the name member, and pass that expression to sizeof. It works because sizeof doesn't attempt to evaluate the pointer expression, it just uses it to compute the size.

Answer 6

You might wanna read this, as it discusses the very same issue and provides all the options mentioned in this thread, and a little more.

Answer 7

struct record
{
    static const int kMaxNameChars=50;
    int id;
    TCHAR name [kMaxNameChars];
};


sizeof(TCHAR)*record::kMaxNameChars //"sizeof(record.name)"
//record::kMaxNameChars sufficient for many purposes.

Portable, perfectly safe and IMO being explicit about raw array length is good practice.

(edit: you might have to macro it in C, if the compiler gets upset about variable array lengths. if you do, consider defining a static const int to the value of the macro anyway!)