If there is no sizeof operator in C

[sumanish]

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.

 

[nbaztec]

sizeof operator is present in C.

 

[Vyom]

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.

 

[tkin]

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

 

[Liverpool_fan]

sizeof - Wikipedia, the free encyclopedia

 

[Tachy1]

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

 

[audiophilic]

It will greatly depend on the compiler you have

 

[rijinpk1]

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.

 

[krazylearner]

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

 

[nbaztec]

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.

        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.