How do I iterate vector - c++

Question

I have such a code
#include <iostream>
#include <vector>
#include <string>
using namespace std;
void removeFirstFromVec(vector<int> & vecLink) {
// remore first element from some vector
vecLink.erase(vecLink.begin() + 0);
}
int main()
{
vector<int> myVec;
myVec.push_back(1);
myVec.push_back(2);
myVec.push_back(3);
cout << "Before removal\n";
for (auto & i : myVec) {
cout << myVec[i-1] << endl;
}
removeFirstFromVec(myVec);
cout << "After removal\n";
for (auto & i : myVec) { // starts with 2
cout << myVec[i-1] << endl;
}
return 0;
}
But on the place where I put a comment it starts with 2 instead of 0 and it causes an error. What I've done wrong or is there a way to use something brief like auto & i : myVec instead of for (int i = 0; i < ...size(); i++) without such an error

Answer 1

In the range-based for loop
for (auto & i : myVec) {
cout << myVec[i-1] << endl;
}
i is the element of the vector, not the index. It should be
for (auto & i : myVec) {
cout << i << endl;
}

Answer 2

Iteration returns the items themselves, not indices.
vector<int> vec { 5, 3, 2 };
for (int i: vec) {
cout << i << endl;
}
Will output this:
5
3
2

Answer 3

If you do the following:
Use templated-function to accept any type for the vector.
Use direct initialization method rather than push_back() to avoid huge number of lines.
Braces for single syntax is optional. Avoid it to reduce confusion during coding complex programs.
Then you can write the same code in a better manner (don't forget the comments):
#include <iostream>
#include <vector>
using namespace std;
// defining a templated-function
template<class T>
void removeFirstFromVec(vector<T>& vecLink) {
vecLink.erase(vecLink.begin());
}
int main(void) {
// initializing with direct-initialization method
vector<int> myVec {1, 2, 3};
cout << "Before removal\n";
// no braces
for (auto& i : myVec)
// 'i' isn't an index here, it's an element
cout << i << endl;
removeFirstFromVec(myVec);
cout << "After removal\n";
// no braces
for (auto& i : myVec)
cout << i << endl;
return 0;
}