error: scalar object 'v' requires one element in initializer

Question

Hi,

I have some simple code shown below, saved into a file (let us say stock_portfolio.cxx).

I am trying to compile this as: g++ stock_portfolio.cxx

but I get the following error during compiling stage:

error: scalar object 'v' requires one element in initializer

The gcc version I have is: gcc (GCC) 4.1.2 20080704 (Red Hat 4.1.2-52)

#include<iostream>
#include<vector>
int main() {
 std::vector<int>v = {1,2,3,4,5,6};
 //std::vector<string> four_star_stocks;
 for(int i = 0; i < v.size(); ++i){
    std::cout << "Stock S&P: " << v[i] << "\n";
 }
 std::cout << "========================" << "\n";
 std::cout << "Totals   : " << v.size() << "\n";
 return 0;
}

4.1.2 ? Um... does anyone know off the top of their head when gcc added support for initializer-lists ? — WhozCraig
– WhozCraig, Commented Oct 9, 2014 at 21:34
@cdhowie yeah, i kinda figured 4.1.2 was a little ancient for that feature. — WhozCraig
– WhozCraig, Commented Oct 9, 2014 at 21:36
Note that prior to C++11 you can use boost::assign for this task: std::vector<int> v = boost::assign::list_of(1)(2)(3)(4)(5)(6);. — cdhowie
– cdhowie, Commented Oct 9, 2014 at 21:37

EJ Googins · Accepted Answer · 2014-10-09 21:38:52Z

2

list-initialization was only introduced to C++ in C++11. gcc version 4.1 doesn't support C++11 (see https://gcc.gnu.org/projects/cxx0x.html)

It's unclear to me if you're question is asking for a suggested solution/fix or an explanation of why your code will not compile.

answered Oct 9, 2014 at 21:38

EJ Googins

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Comments

Community · Accepted Answer · 2017-05-23 12:22:23Z

Initialize your vector in a loop, like this:

for(int i = 1; i <= 6; ++i)
  v.push_back(i);

As cdhowie suggests, your version of gcc doesn't support the initializer list (you need at least g++ version 4.4, source). In case you get a newer one (adding the flag -std=c++0x or -std=gnu++0x), then you could see the following:

std::vector<int> v = {1,2,3,4,5,6};

or

If you want to do that with an initializer list, then you should use std::initializer_list like this:

#include <iostream>
#include <vector>
#include <initializer_list>

template <class T>
struct S {
    std::vector<T> v;
    S(std::initializer_list<T> l) : v(l) {
         std::cout << "constructed with a " << l.size() << "-element list\n";
    }
    void append(std::initializer_list<T> l) {
        v.insert(v.end(), l.begin(), l.end());
    }
    std::pair<const T*, std::size_t> c_arr() const {
        return {&v[0], v.size()};  // list-initialization in return statement
                                   // this is NOT a use of std::initializer_list
    }
};

template <typename T>
void templated_fn(T) {}

int main()
{
    S<int> s = {1, 2, 3, 4, 5}; // direct list-initialization
    s.append({6, 7, 8});      // list-initialization in function call

    std::cout << "The vector size is now " << s.c_arr().second << " ints:\n";

    for (auto n : s.v) std::cout << ' ' << n;

    std::cout << '\n';

    std::cout << "range-for over brace-init-list: \n";

    for (int x : {-1, -2, -3}) // the rule for auto makes this ranged for work
        std::cout << x << ' ';
    std::cout << '\n';

    auto al = {10, 11, 12};   // special rule for auto

    std::cout << "The list bound to auto has size() = " << al.size() << '\n';

//    templated_fn({1, 2, 3}); // compiler error! "{1, 2, 3}" is not an expression,
                             // it has no type, and so T cannot be deduced
    templated_fn<std::initializer_list<int>>({1, 2, 3}); // OK
    templated_fn<std::vector<int>>({1, 2, 3});           // also OK
}

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