Initializing a vector using string literals

Question

What's correct behavior for following code ?

#include <vector>
#include <iostream>

int main() 
{
  std::vector<char> v = { "y", "z" };  
  std::cout << v[0];

  return 0;
}

This is accepted by Clang but not by GCC and VC++

Is this not an undefined behavior ?

It's undefined behaviour. Related: http://stackoverflow.com/questions/24112281/c11-initializer-list-fails-but-only-on-lists-of-length-2 — chris, Nov 10 '14 at 05:23
AHA! I finally found it! http://cppquiz.org/quiz/question/129 — chris, Nov 10 '14 at 05:53
What is the point of using vector with a single string literal? Can't you use a char literal? — Germán Diago, Nov 10 '14 at 09:04

score 11 · Accepted Answer · edited May 23 '17 at 12:07

After digging a bit into standards, I found following :

Here, I am trying to initialize vector<char> using two string literals, not two chars. using vector(initializer_list<T>). In this case, vector(initializer_list<char>).

But the type of a string literal is "array of n const char", so the initializer-list constructor is not a match.

This doesn't result in compiler error, since the compiler is able to find another constructor that matches

§13.3.1.7¶1 explains the rules :

"When objects of non-aggregate class type T are list-initialized, overload resolution selects the constructor in two phases:

— Initially, the candidate functions are the initializer-list constructors of the class T and the argument list consists of the initializer list as a single argument [which we have seen didn't match].

— If no viable initializer-list constructor is found, overload resolution is performed again, where the candidate functions are all the constructors of the class T and the argument list consists of the elements of the initializer list."

And the match in this case is :

template <class InputIterator> vector(InputIterator first, InputIterator last)

The type of InputIterator has no info of T in the vector<T>. So even if I'm initializing a vector<char>, the two arguments can be of arbitrary type. The only requirement is that they stick to InputIterator property, which const char[] happens to do.

The constructor believes it has been passed two iterators to the same sequence, but it has actually been passed iterators to two completely different sequences, "y" and "z".

So the result of this program is undefined .

^{Thanks to chris's comment
This is post is exactly same as he mentioned over there. See this}

Initializing a vector using string literals

1 Answers1