does C remember the no of elements in the Array??

[L.O.K.]

int a[99];

there are 99 (int)s memory allocated.

why we use sizeof( a ) , it can output 198 ?

why it knows ? where does it store the no of elements?

[Hammer]

sizeof(a) gives you the size of the whole array, which is made up of sizeof(int) * 99.
It doesn't store the number of elements, but you can work it out with some simple maths:
sizeof(a) / sizeof(a[0])

But be careful if you pass "a" to a function, 'cos sizeof might not do as you expect.

[L.O.K.]

O~~ thx..

my Eng. is poor..

i want to ask how C knows that the no. of element is 99 ??

and ,, can we get 99 in other method of sizeof(a)/sizeof(a[0]) ??

just want to know..

[Hammer]

#include <stdio.h>
 
 int main(void)
 {
   int a[99];
   
   printf ("sizeof(a)					  :%lu\n", sizeof(a));
   printf ("sizeof(a[0])				   :%lu\n", sizeof(a[0]));
   printf ("sizeof(int) (same as previous) :%lu\n", sizeof(int));  
   printf ("sizeof(a) / sizeof(a[0])	   :%lu\n", sizeof(a) / sizeof(a[0]));
   printf ("396 / 4					    :%lu\n", 396 / 4);
   
   return(0);
 }
 
 /*
 
  My Output:
  
  sizeof(a)					  :396
  sizeof(a[0])				   :4  
  sizeof(int) (same as previous) :4  
  sizeof(a) / sizeof(a[0])	   :99 
  396 / 4					    :99
  
  */

(excuse the poor layout, my browser seems to be doing strange things!)

[Brain Cell]

the size of the type INT in your computer is 2 bytes. So the size an array of 99 elements , 2 bytes for each element , is 198.

and ,, can we get 99 in other method of sizeof(a)/sizeof(a[0]) ??

yes , try this :

sizeof(a) / 2

whereas 2 represent the size of int in your computer. Or more accurate\portable code :

sizeof(a) / sizeof(int)

hammer used "sizeof(a) / sizeof(a[0])" where 'sizeof(a[0])' means the size of one element (the 0 can be replaced with any number within the array range), and since its an INT array , the sizeof(a[0]) would return 2 in your computer and 4 in hammer's , because an INT is either 2 or 4 bytes big (depends on how the computer treats it).

hope this clears things up

[quzah]

But again, as Hammer stated, this method will not work unless you are in the original declaring scope of the array:

void foo( int bar[99] )
{
    printf("The size of the array in foo: %d\n", sizeof( bar ) / sizeof( bar[0] ) );
}

int main( void )
{
    int baz[99];
    printf("The size of the array in declared scope: %d\n",
        sizeof( baz ) / sizeof( baz[0] ) );

    foo( baz );

    return 0;
}

Thus, you can only get the size of an array in the scope in which it was declared.

Quzah.

[L.O.K.]

char a[99];


sizeof(a); // <<--- How does "sizeof" work to know the no. of elements is 99 ?? I am wondering if there is a place in the memory to store the 99 ??

[Thantos]

sizeof is a compile time operation. So when the compiler sees

char a[99];

it adds a to its symbol table and keeps track of it's type and its size.

[L.O.K.]

OK , i understood now. thank .'