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*?

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. — Mooing Duck
– Mooing Duck, Commented Aug 30, 2012 at 0:23

Mankarse · Accepted Answer · 2012-08-30 00:21:34Z

8

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.

answered Aug 30, 2012 at 0:21

Mankarse

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Comments

Grzegorz · Accepted Answer · 2012-08-30 00:22:43Z

1

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.

answered Aug 30, 2012 at 0:22

Grzegorz

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