convert int to pointer int *ptr position; - c++

Question

i have 5 digits in 1 pointer
int* reversevec(int* ptr, unsigned int Ne){
int* ReverResult;
unsigned int rever=Ne, z;
ReverResult=(int*)malloc(Ne*sizeof(int));
for (int i = 0; i < Ne; ++i)
{
z=*(ptr+rever);
printf("%i ",z);//to be sure z starts from the last number on ptr to the fisrt
rever--;
}
return ReverResult;
}
example
Number of elements(Ne)=5
int* ptr have 5 numbers {1 2 3 4 5}
every time when z is printed i got {5 4 3 2 1}
but i cant save z into *ReverResult
ReverResult=(z+rever);
this line is what i tried to put into cicle for to save z and position into int pointer ReverResult but i cant convert int to int*;

Answer 1

There are many problems here
z is a local variable int.
its address will not be useful to return, because it will be out of scope.
returning an offset from its address is even worse, since that is a totally unrelated place in memory.
you also have an off-by-one error. imagine Number elements is one. You will then try to view ptr+1 instead of ptr+0.
you've also tagged this c++ but are writing c style code.
to answer your primary question, rather than writing ReverResult=(z+rever) one could write *(ReverResult + rever - 1) = *(ptr + i)

Answer 2

The other way round, you need to dereference your pointer to be able to assign. After all, you are easier off with pure pointer arithmetics:
int* result = malloc(sizeof(*ptr) * ne);
// let's just have to pointers running through the two arrays:
for(int* p = ptr + ne, *r = result; p-- != ptr; ++r)
{
*r = *p;
// ^ ^ this is the important part: dereference both pointers
}
If you still prefer indices you can use operator[]:
--ne; // so that we don't have to subtract within the loop again and again
for(size_t index = 0; index <= ne; ++index)
{
result[index] = ptr[ne - index];
// ^^^^^^^ again: dereferencing, this time with offset
}
array[index] is entirely equivalent to *(array + index) – and actually it, too, to index[array] (e. g. 7[array]), which is often used to fool the inexperienced...