int main()
{
int a[]={1,2,3,4,5,6,7,8,9,0};
printf("a = %u , &a = %u\n",a,&a);
printf("a+1 = %u , &a+1 = %u\n",a+1,&a+1);
}
how a and &a are internally interpreted?
Both statements print out addresses and are probably meant to explain pointer arithmetic.
a
and &a
are NOT the same, they have different types, but hold the same memory address.
&a
is of type int (*)[10]
(which acts like a pointer to an array)a
is of type int [10]
(which acts like a pointer to a single element)
So when you add 1 keep those types in mind. The pointer will be offset by the size of the type that the address contains. a+1
offsets by the size of int, i.e. to the second element in the array. &a+1
offsets completely past the whole array.
Well, a is the address of the first element of the array, and &a is the address of the array, but obviously they both have the same address.
However when you add (or subtract) a number from a pointer the compiler takes the size of the data into consideration thus in your case (assuming the size of int is 4 bytes) a+1 will be bigger than a by 4 because you move the pointer one integer ahead, but &a+1 would be bigger by 40 because you more the pointer one ARRAY OF 10 INTEGERS ahead.
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