Question

#include <iostream>
#include <typeinfo>

int main()
{
    const char a[] = "hello world";
    const char * p = "hello world";
    auto x = "hello world";

    if (typeid(x) == typeid(a))
        std::cout << "It's an array!\n";

    else if (typeid(x) == typeid(p))
        std::cout << "It's a pointer!\n";   // this is printed

    else
        std::cout << "It's Superman!\n";
}

Why is x deduced to be a pointer when string literals are actually arrays?

A narrow string literal has type "array of n const char" [2.14.5 String Literals [lex.string] §8]

Answer 1

The feature auto is based on template argument deduction and template argument deduction behaves the same, specifically according to §14.8.2.1/2 (C++11 standard):

• If P is not a reference type
  • If A is an array type, the pointer type produced by the array-to-pointer conversion is used in place of A for type deduction

If you want the type of the expression x to be an array type, just add & after auto:

auto& x = "Hello world!";

Then, the auto placeholder will be deduced to be const char[13]. This is also similar to function templates taking a reference as parameter. Just to avoid any confusion: The declared type of x will be reference-to-array.