I was reading more about arrays vs pointers in C and wrote the following program.
#include <stdio.h>
int arr[10] = { } ;
typedef int (*type)[10] ;
int main()
{
type val = &arr ;
printf("Size is %lu\n", sizeof(val)) ;
printf("Size of int is %lu\n", sizeof(int)) ;
}
If, I execute this program, then sizeof(val) is given to be 8 and sizeof(int) is given to be 4.
If val is a pointer to the array with 10 elements, shouldn't it's size be 40. Why is the sizeof(val) 8 ?
If val is a pointer to the array...
Yes, it is, and sizeof(val) produces the size for the "pointer to the array", not the array itself.
...shouldn't it's size be 40.?
No, sizeof(val) calculates the size of the operand, the "pointer" here. In your platform, the size of a pointer seems to be 64 bits, i.e., 8 bytes. So, it gives 8.
Also, as I mentioned, use %zu to print size_t, the type produced by sizeof operator.
int(*)[10] means a pointer to an array of 10 ints while int* is just a pointer to an int.sizeof(int) in your question, that is not a pointer. Did you mean sizeof(int *)??First of all this initialization of an array
int arr[10] = { } ;
is invalid in C. You may not use emplty braces in C (in C++ they are allowed). You have to write
int arr[10] = { 0 } ;
In C the corresponding initializer is defined the following way
initializer:
{ initializer-list }
{ initializer-list , }
while in C++
braced-init-list:
{ initializer-list ,opt }
{ }
As for this statement
printf("Size is %lu\n", sizeof(val)) ;
then val is a pointer because it has type type defined like
typedef int (*type)[10] ;
Change this statement to
printf( "Size is %zu\n", sizeof( *val ) );
if you want to get the size of the object (that is of the array) pointed to by the pointer.
c99 also? I am sure ISO C forbids it, but not very sure about c99, can you confirm?
sizeof returns the size of the pointer itself. In 64bit system it is 8 bytes. Pointers have no knowledge of the size of buffer they points to.
A pointer is a distinct data type and always has a fixd size. Think of a pointer as a sign that points to the actual data, and the sign always the same size - regardless if it points to a single character or a larger array.
val is pointer and its size is equivalent to address bus size of system.
so, sizeof(val) or sizeof(anyPointer) will give you 8 as output.
if you want 40 then try sizeof(arr).
Pointer to an array is simple pointer where you can write val++ while array name is constant pointer you cannot write arr++.
you can also check this link
Most probably you are in a 64 bit PC where memory address need 64 bits or 8 byte space for representation.
Now the pointer actually holds the memory address, whereas it may hold an address of int or maybe an address of int[], doesn't matter.
So, when you execute,
printf("Size is %lu\n", sizeof(val)) ;
it shows 8 as because the pointer holds address which need 64 bits space.
And
printf("Size of int is %lu\n", sizeof(int)) ;
this line just print the size of int which is 4 bytes.
%zuto printsizeof.