Variadic function dosen't accept string objects

Question

Reading about variadic functions, I found a sum function which accepts any number of any numeric type and calculate sum of them.

Having templated nature of this function, I expected it accepts string objects since operator + is defined for strings.

#include <iostream>
#include <string>
#include <type_traits>
#include <utility>

using namespace std;

template <typename T> T sum(T && x)
{
    return std::forward<T>(x);
}

template <typename T, typename ...Args>
typename std::common_type<T, Args...>::type sum(T && x, Args &&... args)
{
    return std::forward<T>(x) + sum(std::forward<Args>(args)...);
}

int main()
{
    auto y = sum(1, 2, 4.5); // OK
    cout << y << endl;

    auto x = sum("Hello!", "World"); // Makes error
    cout << x << endl;

    return 0;
}

Error:

invalid operands of types 'const char [7]' and 'const char [6]' to binary 'operator+'

I expected it concatenates Hello! and World and prints out Hello!World. What is the problem?

const char* doesn't have overloaded operator+, like it says in the error. Pretty clear if you ask me. — Rapptz
– Rapptz, Commented Feb 24, 2013 at 21:17

Andy Prowl · Accepted Answer · 2013-02-24 21:33:33Z

2

String literals are not std::string objects. No operator + is defined for arrays of characters.

As your compiler is telling you, "Hello!" has type const char[7], while "World" has type const char[6]. Try declaring two variables of those types and taking their sum:

int main()
{
    char const a[7] = "Hello!";
    char const b[6] = "World";
    (a + b);
}

And the compiler will show you a similar error:

error: invalid operands of types 'const char [7]' and 
       'const char [6]' to binary 'operator+'

To make your code work, wrap at least one of the two string literals into an std::string object (two corresponding overloads of operator + exist for std::string objects):

auto x = sum(std::string("Hello!") + "World");

or:

auto x = sum("Hello!" + std::string("World"));

Of course, you can also wrap both arguments, but that's unnecessary.

edited Feb 24, 2013 at 21:33

answered Feb 24, 2013 at 21:18

Andy Prowl

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3 Comments

masoud Over a year ago

auto x = sum(string("Hello!"), string("World")); !

Andy Prowl Over a year ago

@MM.: Yes. That's because you wrap the literals in string objects

masoud Over a year ago

It's better I go sleep before asking worse questions!

Xeo · Accepted Answer · 2013-02-24 21:21:36Z

2

The underlying problem isn't with variadic templates, but with your expectations - string literals, like "hello" are not of type std::string. They're of type char const[N] where N is the number of characters + 1. If you actually construct a string from them (or even just from the first one), it works as expected:

// snip

int main()
{
    auto y = sum(1, 2, 4.5); // OK
    cout << y << endl;

    auto x = sum(std::string("Hello!"), "World"); // OK
    cout << x << endl;

    return 0;
}

Live example.

answered Feb 24, 2013 at 21:21

Xeo

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