Thread: Sizeof

Code:

#include <stdio.h>

size_t getSize( float *ptr ); /* prototype */

int main( void )
{
   int array[ 20 ]; /* create array */

   printf( "The number of bytes in the array is %d"
           "\nThe number of bytes returned by getSize is %d\n",
           sizeof( array ), getSize( array );

   return 0; /* indicates successful termination */

} /* end main */g

/* return size of ptr */
size_t getSize( float *ptr )
{
   return sizeof( ptr );

} /* end function getSize */

The sizeof(array) will get 80 while getSize(array) will get 4. Why it's 4. Does this means the pointer point to the first array. Example: array[0].

An int (on most architectures) takes up four bytes of memory. You're allocating 20 ints, hence, 80 bytes.

The getSize function returns the size of the pointer, which is usually 4 or 8 bytes, and used to reference the location in memory that the first element is stored.

ok thank you. Understand.

Is pointer 4 bytes on 32 bit machine and 8 on 64? or is that irrelevant

Originally Posted by camel-man

Is pointer 4 bytes on 32 bit machine and 8 on 64? or is that irrelevant

That's generally the case, though the "bits" of a machine typically refer to the word (data) size, not the "address size".