If there is no sizeof operator in C

S

I do not have much knowledge about compilers and all.

I have one simple question about C programming.

If there is no sizeof operator in C then does those pointer arithmetic work or not ?

I am very much confused about it. :-?

Please somebody help.

sizeof operator is present in C.

sumanish said:
If there is no sizeof operator in C then does those pointer arithmetic work or not ?

I think you need to read some books. Some good books. Non-Indian authors.

I learned about this operator when learning C, when doing dynamic memory allocation, read up on malloc, calloc.

sizeof - Wikipedia, the free encyclopedia

T

yes there is ....if u r having probs using calloc,malloc etc maybe u forgot to use type casting..

It will greatly depend on the compiler you have

sizeof() is put by the compiler itself when running pointer arithmatic. Without that, pointer arithmatic wont takeplace. That is what I understand as of now.

K

hii there i learned this trick back in my college days to find the size of a variable without the use of sizeof() operator .

#define find_size(x) (((char *)(&x+1))-((char *)(&x)))

int main(){
int i;
char c;
double d;
printf("%d\n",find_size(i));
printf("%d\n",find_size(c));
printf("%d",find_size(d));
return 0;
}

It works in almost all cases .Although sizeof () is a operator i think in assembly code it comes down to some piece of more basic code .

don't ask how to find size() if we don't have # define

That i don't know

krazylearner said:
hii there i learned this trick back in my college days to find the size of a variable without the use of sizeof() operator .

It works in almost all cases .Although sizeof () is a operator i think in assembly code it comes down to some piece of more basic code .

don't ask how to find size() if we don't have # define
That i don't know

As cool and hacky as the trick may sound, it's bad code. And not because it does what sizeof() does, but because it is misleading and counter-intuitive. One should take some time out to understand what exactly is happening:

A pointer is nothing but a location to memory, and on a x86 systems it holds a 32-bit address. So effectively it is of type long (4 bytes) = int on x86 architecture.

Code:

        int i = 1337;
	printf("i(%d) is located at %d, next location is %d\n", i, &i, &i+1); 
	printf("Number of memory locations: %d - %d = %d\n", &i+1, &i, &i+1 - &i);
	printf("Number of bytes           : %d - %d = %d\n", &i+1, &i, (long)(&i+1) - (long)&i);

     /**
        * Output
	* i(1337) is located at 1834544, next location is 1834548
	* Number of memory locations: 1834548 - 1834544 = 1
	* Number of bytes           : 1834548 - 1834544 = 4
        */

So without typecasting, the difference between 2 pointers yields the no. of elements or memory locations. But once you typecast the pointer to a long, it is treated as a interger type memory location upon which you can perform the usual arithmetic operations. And what better way to get the number of bytes that to get the difference between 2 consecutive addresses.

So you can ditch that hacky macro, which I'm sure originated from some Indian authored book.