Remove an array element and shift the remaining ones

You just need to overwrite what you're deleting with the next value in the array, propagate that change, and then keep in mind where the new end is:

int array[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};

// delete 3 (index 2)
for (int i = 2; i < 8; ++i)
    array[i] = array[i + 1]; // copy next element left

Now your array is {1, 2, 4, 5, 6, 7, 8, 9, 9}. You cannot delete the extra 9 since this is a statically-sized array, you just have to ignore it. This can be done with std::copy:

std::copy(array + 3, // copy everything starting here
          array + 9, // and ending here, not including it,
          array + 2) // to this destination

In C++11, use can use std::move (the algorithm overload, not the utility overload) instead.

More generally, use std::remove to remove elements matching a value:

// remove *all* 3's, return new ending (remaining elements unspecified)
auto arrayEnd = std::remove(std::begin(array), std::end(array), 3);

Even more generally, there is std::remove_if.

Note that the use of std::vector<int> may be more appropriate here, as its a "true" dynamically-allocated resizing array. (In the sense that asking for its size() reflects removed elements.)

You can use memmove(), but you have to keep track of the array size yourself:

size_t array_size = 5;
int array[5] = {1, 2, 3, 4, 5};

// delete element at index 2
memmove(array + 2, array + 3, (array_size - 2 - 1) * sizeof(int));
array_size--;

In C++, though, it would be better to use a std::vector:

std::vector<int> array;
// initialize array...

// delete element at index 2
array.erase(array.begin() + 2);