size of a datatype without using sizeof

Question

Hi all
I have a datatype say X and I want to know its size without declaring a variable or pointer of that type and of course without using sizeof operator.
Is this possible. I thought of using standard header files which contain size and range of datatypes but that doesn't work with user defined datatype.
Any help will be appreciated.

Thanx

Answer 1

Look into the compiler sources. You will get :

• the size of standard data types.
• the rules for padding of structs

and from this, the expected size of anything.

If you could at least allocate space for the variable, and fill some sentinel value into it, you could change it bit by bit, and see if the value changes, but this still would not tell you any information about padding.

Answer 2

Look, sizeof is the language facility for this. The only one, so it is the only portable way to achieve this.

For some special cases you could generate un-portable code that used some other heuristic to understand the size of particular objects[*] (probably by making them keep track of their own size), but you'd have to do all the bookkeeping yourself.

[*] Objects in a very general sense rather than the OOP sense.

Answer 3

The possibility of padding prevent all hopes without the knowledge of the rules used for introducing it. And those are implementation dependent.

Answer 4

You could puzzle it out by reading the ABI for your particular processor, which explains how structures are laid out in memory. It's potentially different for each processor. But unless you're writing a compiler it's surprising you don't want to just use sizeof, which is the One Right Way to solve this problem.

Answer 5

To my mind, this fits into the category of "how do I add two ints without using ++, += or + ?". It's a waste of time. You can try and avoid the monsters of undefined behaviour by doing something like this.

size_t size = (size_t)(1 + ((X*)0));

Note that I don't declare a variable of type or pointer to X.

Answer 6

Well, I am an amateur..but I tried out this problem and I got the right answer without using sizeof. Hope this helps.. I am trying to find the size of an integer.

int *a,*s, v=10;

a=&v;

s=a;

a++;

int intsize=(int)a-(int)s;

printf("%d",intsize);

Answer 7

The correct answer to this interview question is "Why would I want to do that, when sizeof() does that for me, and is the only portable method of doing so?"

Answer 8

int a; printf("%u\n",(int)(&a+1)-(int)(&a));