Odd behaviour of operator+()

[Juha Nieminen]

Consider this:

#include <iostream>
#include <string>

std::string operator+(const char* a, const std::string& b)
{
return std::string(a)+b;
}

int main()
{
std::string s = "abc" + "def";
std::cout << s << std::endl;
}

With gcc 3.3.5 it gives the rather strange error:

test.cc:11: error: invalid operands of types `const char[4]' and `const
char[4]
' to binary `operator+'

If I substitute the offending line with this:

std::string s = "abc" + std::string("def");

or even with this:

std::string s = operator+("abc", "def");

then it works just ok. However, as given above it doesn't. How come?
(References to the C++ standard preferred.)

[Thorsten Kiefer]

Hi,
there is no instance of operator+(const char *,const char *).
try this :

std::string operator+(const char* a, const char *b)
{
return std::string(a)+std::string(b);
}

I don't know why "std::string s = operator+("abc", "def");" works !?!
Regards
Thorsten

[Roland Pibinger]

On Sat, 24 Jun 2006 21:25:45 +0300, Juha Nieminen
<no****@thanks.invalid> wrote:

Consider this:

#include <iostream>
#include <string>

std::string operator+(const char* a, const std::string& b)

BTW, this function is already implemented by std::(basic_)string.

[Alf P. Steinbach]

* Juha Nieminen:

Consider this:

#include <iostream>
#include <string>

std::string operator+(const char* a, const std::string& b)
{
return std::string(a)+b;
}

int main()
{
std::string s = "abc" + "def";
std::cout << s << std::endl;
}

You can overload an operator on class or enum type. You can not
overload the built-in operators. The usage

"abc" + "def"

invokes the built-in operator+, with invalid arguments.

Ignore the replies that indicate you can overload the built-in operator.

You can't.

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