Why does C++ allow a char array as an argument when it's expecting a string?

Question

I have the following code:

#include <iostream>
#include <string>

using namespace std;


string combine(string a, string b, string c);

int main() {

    char name[10]   = {'J','O','H','N','\0'};
    string age      = "24";
    string location = "United Kingdom";


    cout << combine(name,age,location);

    return 0;

}

string combine(string a, string b, string c) {
    return a + b + c;
}

This compiles fine with no warnings or errors despite the combine function expecting a string and receiving a char array, is this because a string is stored as a char array?

char name[10] = {'J','O','H','N','\0'}; can be written much more simply as char name[10] = "JOHN";. — Pete Becker
– Pete Becker, Commented Nov 30, 2013 at 11:36

Sarfaraz Nawaz · Accepted Answer · 2013-11-30 08:36:08Z

5

Why does C++ allow a char array as an argument when it's expecting a string?

Because std::string has such a conversion constructor, which supports implicit conversion of char const* into std::string object.

This is the constructor which is responsible for this conversion:

basic_string( const CharT* s, const Allocator& alloc = Allocator());

Have a look at the documentation and other constructors.

edited Nov 30, 2013 at 8:36

answered Nov 30, 2013 at 8:31

Sarfaraz Nawaz

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Comments

john · Accepted Answer · 2013-11-30 08:32:17Z

1

It's because there is an automatic conversion from a char array to a string.

string has a constructor like this (simplified)

class string
{
public:
    string(const char* s);
    ...
};

This constructor can be called automatically, so your code is equivalent to this

cout << combine(string(name),age,location);

answered Nov 30, 2013 at 8:32

john

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