In this code,
int a[] = {1, 2, 3, 4, 5}; printf("a = %p, &a = %p\n", a, &a); same address for a and &a is printed. As I know, a is a const pointer to 0th element of array. Why address of a and contents of it are equal?
1 Answer
Why address of a and contents of it are equal?
They are not.
In most of the cases, a variable of type array decays to a pointer to the first element. Thus
printf("1. %p, 2. %p", (void*)a, (void *)&a[0]); will print same values.
That said, the address of an array, is the same as the address of the first element of the array, thus
printf("1. %p, 2. %p", (void*)a, (void*)&a); also prints the same value. However, remember, they are not of the same type.
a, which is same as&a[0]in this case, is of typeint *&a, is of typeint *[5], i.e, a pointer to an array of 5ints.
2 Comments
Yousef Rashidi
How a is stored in memory? Any pointer variable, is 4 (or 8) byte in memory, with value = address of variable that pointer points to. Is it incorrect about array name?
Antti Haapala
@yousefrashidi it is not stored in memory at all. C.f. "How is the address of
int a stored in memory?! Doesn't every int a actually need 12 bytes of memory, 4 for the a and 8 for its address?!" And where is this address stored?" Pointers all way down...
%pspecification then do not forget to cast the argument tovoid *:printf("a = %p, &a = %p\n", (void *)a, (void *)&a).ais not a const pointer; it is an array. Have a look at the duplicates.