C++: passing a string-literal of Type const char* to a string-parameter

Question

i'm new to c++ and have a lack of understanding why this code works well:

string GetString(string promt)
{
    cout << promt << ": ";
    string temp;
    getline(cin, temp); 
    return temp; 
}

int main()
{
    string firstName = GetString("Enter your first name"); 
    string lastName = GetString("Enter your last name");

    cout<< "Your Name is: " << firstName << " " << lastName; 


    cin.ignore(); 
    cin.get(); 

    return 0;
}

String-Literals like "bla" are of Type const char*. At least auto i = "bla"; indicates, that i is of Type "const char*". Why is it possible to pass it to the GetString-Function, because the function expects a string and not a const char*?

Unattended

String literals are of type const char(&)[N], but since C can't copy an array to the local variable, auto makes the local const char* anyway. – Lieselotteliestal

Answer 1

std::string has a converting constructor which takes a char const* and initialised the string with the null terminated string pointed to by the pointer. This constructor is not explicit, so it can be used in implicit conversions.

Ideation

Answer 2

Look at the std::string constructors. What happens is that compiler found a constructor that accepts your const char* and used it to automatically convrt const char* to std::string. BTW I would suggest using const std::string& prompt instead.

Barnum