Display Vector elements in a reverse order - c++

Question

I have two questions but they are interlinked.:
part:a->
I have been trying to display the elements of vector in reverse order. But nothing is working. I have used iterotar like;
for (it=vec.end(); it!=vec.begin(); --it){
// it is iterator not reverse_iterator.
// do work
}
P.S I am not much familiar with iterators. I have used them for the first time today to
display elem in reverse order.
also tried;
for (int i=vec.size(); i!=0; i--){
//display
}
No matter what I do it always display the elem in same order as they are present i.e not in the reverse order.
part_b->
Is there any way that I can store the output of a recursive function direct into a vector. like the code is:
I know this does not work. I have tried but just giving you an idea
what I am upto.
#include <iostream>
using namespace std;
#include "vector"
int func(int num);
vector <int> vec;
int main() {
int num=34;
// I know this would not work. But is there any possibilitiy that
// I can store the output in a vector.
vec = binary(num);
// trying to display the vector.
for (int i=vec.size();i!=0;i--) {
cout<<vec[i]<<" ";
} // not working for reverse display.
} //main.
int func(int num) {
if (num==1) {
//vec.push_back(1);
return 1;
}
else if(num==0) {
//vec.push_back(0);
return 0;
}
else {
//vec.push_back(input%2);
return binary(input/2);
}
} //func.
I hope you do unnderstand the question. if I am able to do the part b the there is no need to reverse the elem of the vector.

Answer 1

The standard solution uses reverse iterators:
for (auto it = v.rbegin(); it != v.rend(); ++it)
{
if (it != v.rbegin()) { std::cout << ' '; }
std::cout << *it;
}
Alternatively, you can use indices, but keep the "reversal" idiom and increment the index:
for (std::size_t i = 0; i != v.size(); ++i)
{
if (i != 0) { std::cout << ' '; }
std::cout << v[v.size() - i - 1];
}
Note that reverse iterators are literally just doing something very similar to the explicit loop under the hood. In particular, the base() member function of a reverse iterator gives you the corresponding ordinary iterator offset by one, just as we have a - 1 in the index computation. (E.g. rbegin() is just end() under the hood, but decrements by one upon dereferencing.)

Answer 2

Use reverse iterators:
for (auto it = vec.crend(); it != vec.crbegin(); ++it) {
std::cout << *it << ' ';
}
std::cout << '\n';

Answer 3

Part A
Assuming you haven't got access to C++11:
vector<int>::const_reverse_iterator it;
for (it=vec.rbegin(); it!=vec.rend(); ++it)
{
// do work
}
Part B
It looks very much like you're trying to display a number in binary. Unfortunately the standard flags on ostream only allow hex, decimal or octal as far as I'm aware, but can I suggest a simpler way of doing this?
#include <bitset>
bitset< sizeof(int) << 3 > b(34);
cout << b << endl;
Which gives:
00000000000000000000000000100010
The rather ugly looking sizeof(int) << 3 is just a way of getting the size of an int in bits to avoid truncation.

Answer 4

I wrote little program which may help. Maybe your vector is empty?
#include <iostream>
#include <vector>
using namespace std;
int main(int argc, char *argv[])
{
vector<int> vec;
vec.insert(vec.begin(), 1, 1);
vec.insert(vec.begin(), 1, 2);
vec.insert(vec.begin(), 1, 3);
vector<int>::iterator i;
for (i = vec.end(); i != vec.begin(); --i)
{
cout << *i;
}
cout << *i;
return 0;
}