Why is the size of the dynamic array b not the same as the array a, even though both have 17 elements ?

Question

https://code.sololearn.com/c1KuxtS6JFPh/?ref=app

Answer 1

Kuldeep Singh because every pointer is 8 bit large so sizeof(b) = 8. But sizeof(b[0]) = 4, because it expects an integer at the pointers [0] position which has 4 bytes. And dividing the value with arrSize -> 2

Answer 2

because b is not really a array it just stores the address of the first int

Answer 3

thank you, Aaron Erhardt and Timon Paßlick

Answer 4

b is not an array, it's a pointer. You divide the pointer size by what it points to.

Answer 5

If it was a 32 bit architecture, you'd get the expected 1.

Answer 6

Aaron Erhardt but why is it printing 2 instead of 1 ? thx

Answer 7

Timon Paßlick or using a long long datatype which is 8 bytes large