How to iterate over a generic vector

Question

I've always found C++ templates inscrutable, and C++ error messages more so. I want to understand it, rather than always feeling confused. Here's the latest bizarre experience:

error: conversion from ‘<unresolved overloaded function type>’ to non-scalar type ‘std::vector<int>::iterator {aka __gnu_cxx::__normal_iterator<int*, std::vector<int> >}’ requested

This doozy comes from the following code:

#include <iostream>
#include <vector>

using namespace std;

template <typename T>
void printVector(const vector<T>& v) {
    for (typename vector<T>::iterator iter = v.begin; iter != v.end; iter++) {
        cout << *iter << endl;
    }
}

int main() {
    vector<int> v{1, 2, 3};
    printVector(v);
    return 0;
}

It has no problem iterating over the vector if I use a C++11 range-based loop. I want to learn how to do it using iterators.

Can someone explain what the error means and how to fix it?

It would also be wonderful if I could get a recommendation of a book that explains templates well.

You need `begin()` and `end()`. Voting to close as trivial typo. — juanchopanza, May 31 '16 at 07:53
When writing template code, it is often (not always) a good idea to write it first non-template for a single concrete case. Then, copy/paste it and make it a template. — Yakk - Adam Nevraumont, Jun 01 '16 at 05:03

score 3 · Accepted Answer · edited May 23 '17 at 12:16

See the error message carefully. It says you're trying to convert a function to iterator. You should call them by adding ().

Change

for (typename vector<T>::iterator iter = v.begin; iter != v.end; iter++) {

to

for (typename vector<T>::iterator iter = v.begin(); iter != v.end(); iter++) {
                                                ~~               ~~

For C++ books, The Definitive C++ Book Guide and List

101010 · Answer 2 · 2016-05-31T08:01:43.970

2

You're missing the parentheses from begin and end and since you're passing the vector by const reference you'll need a const_iterator.

template <typename T>
void printVector(const vector<T>& v) {
    for (typename vector<T>::const_iterator iter = v.begin(); iter != v.end(); iter++) {
                             ^^^^^^^^^^^^^^               ^^                ^^
        cout << *iter << endl;
    }
}

Better yet use a ranged base for:

template <typename T>
void printVector(const vector<T>& v) {
    for(auto &&e : v) {
        cout << e << endl;
    }
}

edited May 31 '16 at 08:01

answered May 31 '16 at 07:56

101010

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What is the advantage of using `auto &&e` over `auto e`? – Michael Anderson May 31 '16 at 08:48
1

@MichaelAnderson It's useful for proxies (e.g., `std::vector`). – 101010 May 31 '16 at 08:52
Why is the bitwise `&&` being used over the standard `&`? I notice in my personal example I'm not getting any difference. – kevin_theinfinityfund Sep 07 '20 at 23:50

score 0 · Answer 3 · answered May 31 '16 at 07:53

std::vector<T>::begin and ::end are functions; you need to call them:

for (typename vector<T>::iterator iter = v.begin(); iter != v.end(); iter++) {
//                                              ^^               ^^
    cout << *iter << endl;
}

Additionally, you are trying to convert a const iterator to a non-const one. You need to use vector<T>::const_iterator, or, better yet, auto.

How to iterate over a generic vector

3 Answers3