I'm experiencing a problem. I'm trying to get the number of elements in an int array, without passing any explicit parameter, just the pointer to the int array. Now my curiosity:
int * set;
printf("-- %d --", sizeof(set)); //4
set=(int *) malloc(n*sizeof(int));
printf("-- %d --", sizeof(set)); //4
Why are the values the same since before the malloc it isn't initialized and after it is. Thanks
UPDATE:
Is there any way to get the length of an int array?
Because sizeof is evaluated at compile time and yields the size of the type of set, a pointer to int.
There is no generic way to measure the size of memory pointed to by a pointer in C, other than the special case that strings are null terminated by convention.
sizeof will yield the size of the pointer (4 bytes on a 32-bit system, 8 bytes on a 64-bit system), not of the memory pointed to.
If you want to track the size of memory allocated, options are:
int if your application will never validly use that value.The item you're measuring, set, is a pointer-to-integer ( int* ).
And a pointer-to-integer is 4-bytes.
sizeof does NOT measure the amount of memory allocated to the pointer.
It only measures the "item" itself (in this case, a pointer).
A pointer is 32 bits or 64 bits.
Meaning 4 or 8 bytes.
int * set;
printf("-- %d --", sizeof(set)); //4
set=(int *) malloc(n*sizeof(int));
printf("-- %d --", n*sizeof(int)); //n*sizeof(int) is the size of your malloc'd memory
hope that helps. in your last statement you were still asking the size of the individual integer pointer which was 4 bytes on your machine. to get the size of your malloced area of memory you need to use the same expression as is inside the malloc function.
and an integer pointer isn't necessarily the same size as an int itself as Mat corrected me on
