Correctly calculating the size of an array in C

Question

int a[3][3] = { {8,7,6},{5,4,3},{2,1,0} };
printf("sizeof(a[1]+1): %zu\n", sizeof(a[1]+1));
printf("sizeof(a+1): %zu\n", sizeof(a+1));

i'm trying to figure out why does the program output is:

sizeof(a[1]+1): 8

sizeof(a+1): 8

i expected the output to be

sizeof(a[1]+1): 24  //sizeof(a[1]) + sizeof(a[1]) (12 + 12)

sizeof(a+1): 72     //sizeof(a) + sizeof(a) (36 + 36)

You're getting back the size of a pointer in both expressions. But it isn't clear to me what size you're actually trying to get, or why you think that adding 1 to the expression will double its size. — interjay, Jan 31 '23 at 11:43

Vlad from Moscow · Answer 1 · 2023-01-31T13:06:24.437

In this expression a[1]+1 there is used the pointer arithmetic.

The sub-expression a[1] having the array type int[3] is implicitly converted to pointer to the array element type that is to pointer of the type int *. So the whole expression has the type int *. And this call of printf

printf("sizeof(a[1]+1): %zu\n", sizeof(a[1]+1));

outputs the size of a pointer of the type int *.

In this expression a+1 where there is also used the pointer arithmetic the array designator a having the array type int[3][3] is implicitly converted to pointer to its element type int ( * )[3]. And the full expression has this type.

So this call of printf

printf("sizeof(a+1): %zu\n", sizeof(a+1));

also outputs the size of a pointer.

If you need to get the size of the whole array you can write

printf("sizeof(a): %zu\n", sizeof(a));

If you want to output the size of its elements you can write

printf("sizeof(a[0]): %zu\n", sizeof(a[0]));

or

printf("sizeof(*a): %zu\n", sizeof(*a));

It seems you do not understand the pointer arithmetic as for example in this your statement

sizeof(a+1): 72     //sizeof(a) + sizeof(a) (36 + 36)

This expression

sizeof(a+1)

is not the same as the expression suzeof( a ) + sizeof( a ).

The compiler determines the type of the expression a + 1 that has the type int ( * )[3] as pointed above and yields the size of an object of this type.

I think this quote from the C Standard (6.3.2.1 Lvalues, arrays, and function designators) will be useful for you

3 Except when it is the operand of the sizeof operator or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue. If the array object has register storage class, the behavior is undefined.

So for example sizeof( a ) yields the size of the whole array. In this expression the array designator does not implicitly converted to pointer to the array element type.

However if you will write instead sizeof( a + 0 ) then in this expression a + 0 the array designator will be converted to pointer to the array element type and the operator sizeof yields the size of pointer.

Correctly calculating the size of an array in C

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