Reverse Array with given sentinel value - c++

Question

FYI: I am new to programming.
I have an arraysize of 10 and the sentinel value is 0.
My original array is [1 2 3] ( user input) but my reverse is [0 0 0 0 0 0 0 3 2 1].
I need help to make my reverse array [3 2 1].
Here is my code:
int temp;
for (int i = 0; i < arraysize/2; i++)
{
temp = array[arraysize-1-i];
array[arraysize - i - 1] = array[i];
array[i] = temp;
}
cout << "The reverse array: ";
for (int i = 0; i < arraysize; i++)
cout << array[i]<< ' ';
cout << endl;

Answer 1

Just use the standard library algorithms
auto end = std::find(std::begin(array),std::end(array),0);
std::reverse(std::begin(array),end);
//And if you only want to print the non-zero values:
size_t effectiveArraySize = end - std::begin(array);
If the fixed size array is not part of your requirement, you should put your user data in a vector that automaticaly grows as large as you need, instead of using an array that might turn out to be too small:
std::vector<int> v;
while(true) {
int t;
cin >> t;
if (t == 0) {
break;
}
v.push_back(t);
}
std::reverse(v.begin(),v.end());
That way, you don't have any sentinel values in your array / vector to begin with.

Answer 2

Note: Using the respective functions from the STL (std::reverse and std::find) is better, I was just guessing that you are bound to implement this on your own.
Step one: Write a proper reverse function. One that takes (a pointer to) the beginning as well as (a pointer to) the end of the range that should be reversed.
Step two: Write a function to find (the first position of) your sentinel in an array (given via beginning and end, again)
Step three: Connect the two: Reverse from the beginning to the position of your sentinel.
Example without templates:
void reverse(int * from, int * to) {
while ((to - from) > 1) {
--to;
int temp = *from;
*from = *to;
*to = temp;
++from;
}
}
int const * find(int const * from,
int const * const to,
int const value) {
while ((from != to) && (*from != value)) {
++from;
}
return from;
}
void reverse_until (int * const from,
int * const to,
int const sentinel) {
int const * const position_sentinel = find(from, to, sentinel);
reverse(from, from + (position_sentinel - from));
// return the sentinel position from this function
// if you want only the reversed part
}
Tested with:
int main() {
int test[10];
for (size_t i = 0; i < 10; ++i) {
test [i] = i + 1;
}
reverse_until (test, test + 10, 6);
copy(test, test + 10, ostream_iterator<int>{cout, " "});
return 0;
}
(live here)

Answer 3

You need to find the actual length of the array, before performing the reverse operation, and then use that length for all further operations.
Like this:
int actualArraySize = 0;
while(actualArraySize < arraysize && array[actualArraySize]!=0)
{
actualArraySize++;
}
int temp;
for (int i = 0; i < actualArraySize/2; i++)
{
temp = array[actualArraySize-1-i];
array[actualArraySize - i - 1] = array[i];
array[i] = temp;
}
cout << "The reverse array: ";
for (int i = 0; i < actualArraySize; i++)
cout << array[i]<< ' ';
cout << endl;
Note that, actualArraySize can be less than or equal to arraysize, but, not more than it, because of the condition in while(actualArraySize < arraysize && array[actualArraySize]!=0), which means that stop when either a 0 is found or the size of the array is reached.