Can someone please explain me what is the difference between the two following declarations:
char (*arr_a)[5];
char arr_b[20];
and why:
sizeof (*arr_b) = sizeof (char)
sizeof (*arr_a) = 5*sizeof(char)
2 Answers
char (*arr_a)[5];
declares a pointer to a 5-element array of char.
char arr_b[20];
declares just a 20-element array of char.
So, the output of
sizeof (*arr_a)
should be straight forward -- dereferencing the pointer to an array yields the array and it's size is 5.
The following:
sizeof (*arr_b)
gives 1, because dereferencing the identifier of an array yields the first element of that array, which is of type char.
One thing you need to know to fully understand this is how an array evaluates in an expression:
In most contexts, the array evaluates to a pointer to its first element. This is for example the case when you apply indexing to the array.
a[i]is just synonymous to*(a+i). As the array evaluates to a pointer, this works as expected.There are exceptions, notably
sizeof, which gives you the storage size of the array itself. Also,_Alignofand&don't treat the array as a pointer.
6 Comments
Yaron Scherf
so arr_a isn't just a pointer (of size sizeof(void)) but a pointer to an array and therefore it's size is the size of the array? why isn't sizeof(*arr_a) == sizeof(arr_b)?
Yaron Scherf
so what would i need to write in order to get size: sizeof(XX)==20*sizeof(char) using arr_b?
dbush
@FelixPalmen He actually wrote
*arr_a in the comment. It wasn't escaped properly so it shows as italics. |
arr_a is a pointer to an array of 5 char while arr_b is an array of 20 chars. arr_b is not a pointer unlike arr_a.
sizeof (*arr_b) equals to sizeof (char) because *arr_b is of type char (equivalent to arr_b[0]). For
sizeof (*arr_a) equals to 5*sizeof(char) because *arr_a refers to an array of 5 chars and sizeof returns the size of array which is 5.
4 Comments
Yaron Scherf
but why do their sizes are different? I would expect that the size of arr_a will be either size of pointer (sizeof(void)) or the size of char (like *arr_b)
haccks
@YaronScherf; You are not calculating the size of
arr_a, but *arr_a.dbush
@haccks He actually wrote
*arr_a in the comment. It wasn't escaped properly so it shows as italics.haccks
@dbush; OK. I didn't get that. But still answer is valid.
=!===sizeof(char)is, essentially by definition, always 1.