pointers with numbers in backwards [closed] - c++

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I need to receive an undetermined number of integer numbers in between (0-9). With this numbers, print them forwards and backwards, and then erase the numbers at the corners.
Example:
3 5 1 9 4 6 2 4 4 2 6 4 9 1 5 3
5 1 9 4 6 2 4 4 2 6 4 9 1 5
1 9 4 6 2 4 4 2 6 4 9 1
9 4 6 2 4 4 2 6 4 9
4 6 2 4 4 2 6 4
6 2 4 4 2 6
2 4 4 2
4 4
And here is the code I have so far:
#include <iostream>
using namespace std;
int a;
int p;
int set;
void numberss()
{
for set[](int a=0; a<p; a++)
}
int main()
{
cin >> p;
cin >> a;
const int SIZE = p;
int set[] = {a};
int *numPtr;
numPtr = set;
for (int index = 0; index < SIZE; index++)
{
cout << *numPtr << " ";
numPtr++;
}
for (int index = 0; index < SIZE; index++)
{
numPtr--;
cout << *numPtr << " ";
}
return 0;
}

Answer 1

If we ignore errors, you can read the numbers in one at a time, and form a string for the first line of output. Forming the string will involve appending a reversed copy to the original. Once the string is formed, you can output that string for the first line. Then replace the first number with a space character, and shrink the string from the back by two characters. Keep doing that until you are done.
This works because the numbers are all single digit.
int main (void)
{
int N;
std::string nums;
std::cin >> N;
for (int i = 0, x; i < N; ++i) {
std::cin >> x;
nums += std::to_string(x) + ' ';
}
nums.append(nums.rbegin() + 1, nums.rend());
for (int i = 0; i < N; ++i) {
std::cout << nums << '\n';
nums[2*i] = ' ';
nums.resize(nums.size()-2);
}
}
DEMO

Answer 2

Your code doesn't work because you are not reading all of the numbers from the user's input, you are only reading the count and the 1st number. Also, you are not looping enough times to output the numbers in a triangular fashion, you are only outputting the 1st line of the triangle.
Try this instead:
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int p;
vector<int> set;
cin >> p;
set.resize(p);
for (int i = 0; i < p; ++i)
cin >> set[i];
for (int index = 0; index < p; index++)
{
int *numPtr = &set[index];
for (int i = 0; i < index; ++i)
cout << " ";
for (int i = index; i < p; ++i)
cout << *numPtr++ << " ";
for (int i = index; i < p; i++)
cout << *--numPtr << " ";
cout << endl;
}
return 0;
}
Live Demo
That being said, here is an alternative approach that is more C++-ish and less C-ish, by using iterators instead of pointers, and using STL algorithms. Also, you should always validate user input before using it:
#include <iostream>
#include <vector>
#include <iomanip>
#include <algorithm>
#include <limits>
#include <iterator>
#include <cstdint>
using uint16vec = std::vector<uint16_t>; // there is no operator>> for uint8_t...
int main()
{
size_t count = 0;
std::cin >> count;
uint16vec set;
set.reserve(count);
for (size_t i = 0; i < count; ++i)
{
uint16vec::value_type num;
while (!((std::cin >> num) && (num <= 9)))
{
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
std::cin.clear();
std::cout << "Enter a valid number 0..9!" << std::endl;
}
set.push_back(num);
}
auto begin = set.begin(), end = set.end();
auto rbegin = set.rbegin(), rend = set.rend();
auto out = std::ostream_iterator<uint16vec::value_type>(std::cout, " ");
std::cout << std::setfill(' ');
for (size_t i = 0; i < count; ++i)
{
std::cout << std::setw((i*2)+1);
std::copy(begin++, end, out);
std::copy(rbegin, rend--, out);
std::cout << std::endl;
}
return 0;
}
Live Demo

Answer 3

Give this a shot, mate. I've commented it so you can follow along. It allows for characters other than numbers because that seemingly wasn't a requirement of your assignment, so it's on you to sort and filter those out. It prints as intended, so here's hoping that this is what you're looking for to start off with your search for the right answer.
#include <iostream>
#include <vector>
#include <string>
using namespace std;
void pop_front(vector<char>& _vector)
{
_vector.front() = move(_vector.back());
_vector.pop_back();
}
int main()
{
vector<char> characterList;
string consoleInput;
cin >> consoleInput;
//Initialize by pushing everyting from the console into our vector.
for (char c : consoleInput)
{
characterList.push_back(c);
}
//After adding from the console, now we'll do it in reverse.
for (int i = consoleInput.length() - 1; i > -1; i--)
{
characterList.push_back(consoleInput[i]);
}
//Print where we are currently up to for display purposes.
for (char c : characterList)
{
cout << c;
}
// Newline.
cout << "\n";
//Now let's start to chop it down. It'll take the same iterations as the console input, because we're doubling down.
for (int i = 0; i < consoleInput.length() - 1; i++)
{
characterList.erase(characterList.begin());
characterList.pop_back();
for (char c : characterList)
{
cout << c;
}
cout << "\n";
}
return 0;
}