problem with sizeof operator

Question

As i want to find array size dynamically in function, i used sizeof operator. But i got some unexpected result. here is one demo program to show you, what i want to do.

Answer 1

S is an int *, a pointer to an integer, which is a memory address, which is on your machine twice the size of an integer.

If you want the size of the array (I.e., the number of elements), you can't get that directly in pure C. But since this is a c++ question, there is a way: use a vector, which has a size() method.

Actually, this isn't quite true: within the function that you declare S (and only if it's explicitly initialized at compile time as you do in your example -- even new int[19] doesn't work), the sizeof operator actually does get the correct answer, which is why c++ allows you to do this:

int S[]={1,2,3,2,5,6,25,1,6,21,121,36,1,31,1,31,1,661,6};
vector v(S, S + sizeof(S) / sizeof(int) );

and then you can use v.size() (see these docs).

The template version by Nawaz elsewhere is another excellent suggestion which forces the compiler into carrying the full information about the construction of the c++ array around (again, note that this is all known at compile time, which is why you can be explicit about the size in the argument).