c++ - adding 2 const char*

**joaquim** · 09-30-2014

if:

Code:

string a = "hello" + "hi";

why i can't:

Code:

string &operator+(const char *s1, const char *s2)
{
    string a=(string)s1+s2;
    return a;
}

???
heres the error message:
"'std::string operator+(const char*, const char*)' must have an argument of class or enumerated type"

**jimblumberg** · 09-30-2014

That operator overload must be part of a class, you can't create an overloaded operator without the class.

Jim

**joaquim** · 09-30-2014

Originally Posted by jimblumberg

That operator overload must be part of a class, you can't create an overloaded operator without the class.

Jim

but the operator+() only accept 1 or 2 arguments.. and not 3.
my objective, with my new class is avoiding the casting:

Code:

String d="hello" + "hi";

for works, i must casting the 1st const char*... but i want avoid that.

**jimblumberg** · 09-30-2014

If it is part of a class, please show the complete class definition and implementation.

Jim

**joaquim** · 09-30-2014

Originally Posted by jimblumberg

If it is part of a class, please show the complete class definition and implementation.

Jim

heres the class and some samples:

Code:

class String
{
    private:

    public:
        string b="";
    String(string s="")
    {
        b=s;
    }

    String(const char *s="")
    {
        b=s;
    }

    String& operator=(const string &s)
    {
        b=s;
        return *this;
    }

    String& operator=(const String &s)
    {
        b=s.b;
        return *this;
    }
    operator string()
    {
        return b;
    }

};

//by some reason: these 2 functions must be outside of the class and the 'b' must be public and not private
String& operator+(String &s1, const char *s2)
{
    s1.b += s2;
    return s1;
}

String& operator+(String &s1, String &s2)
{
    s1.b += s2.b;
    return s1;
}

String test="hi";
String test2=" hi";
String test3=test+test2 + "hello";
String test4 =(string) "hi" + "hello";

like you see: the test4 must have the string casting, but i want avoid that.
that's why i was trying do that function:

Code:

string &operator+(const char *s1, const char *s2)
{
    string a=(string)s1+s2;
    return a;
}

isn't a String type return... but you see what i mean

**Elkvis** · 09-30-2014

To overload a binary operator (+, -, etc), at least one side of the operator must be a user-defined type. This is an immutable rule of C++.

**joaquim** · 09-30-2014

Originally Posted by Elkvis

To overload a binary operator (+, -, etc), at least one side of the operator must be a user-defined type. This is an immutable rule of C++.

like you see, i have it:

Code:

String& operator+(String &s1, const char *s2)

but not works with:

Code:

String test4 =(string) "hi" + "hello";

**Elkvis** · 09-30-2014

Originally Posted by joaquim

Code:

String test4 =(string) "hi" + "hello";

You can't just cast a string constant to a string class object. You can use construction syntax:

Code:

string test4 = string("hi") + "hello";

but a C-style cast will not work.

**Elysia** · 09-30-2014

Code:

String& operator+(String &s1, const char *s2)
{
    s1.b += s2;
    return s1;
}
 
String& operator+(String &s1, String &s2)
{
    s1.b += s2.b;
    return s1;
}

operator + shall not modify its arguments.
I imagine you would it pretty strange if:

Code:

int x = 10;
int y = 5;
int z = x + y;

Results in x = 15, y = 5, z = 15.
But that's precisely what your code would do to a string object.
Furthermore, you must never return a reference to a temporay.
So the correct version would be:

Code:

String operator + (const String& s1, const char* s2)
{
    String tmp(s1);
    tmp.b += s2;
    return s1;
}

**antred** · 09-30-2014

Originally Posted by Elysia

So the correct version would be:

Code:

String operator + (const String& s1, const char* s2)
{
    String tmp(s1);
    s1.b += s2;
    return s1;
}

You made a tiny but important typo there. I think you meant:

Code:

tmp.b += s2;
return tmp;

**Elysia** · 09-30-2014

You are correct. Thanks for finding the typo =) I've updated the original post.

**phantomotap** · 09-30-2014

O_o

I realize that some people are going to hate the suggestion, but I always wonder why people prefer concatenation over direct construction.

I get it; I do! That variadic constructor is unusual. You often want the concatenation operators in any event. (You want them for other uses.) The usage makes certain purists cry. In the case of this thread, problems need to be understood and solve in any event.

All of that completely ignored, why don't people offer variadic constructor for such situations!?

Code:

string a("hello", "hi"); // = /* string a = string("hello") + "hi"; */ ~= /* string a = "hello" + "hi"; */

Soma

**King Mir** · 09-30-2014

If you just want to concatenate string literals, you can do so without putting any operator between them. This is useful for multi-line strings.

Code:

const auto str = "hello " 
    "there";

**grumpy** · 10-01-2014

Originally Posted by joaquim

if:

Code:

string a = "hello" + "hi";

Firstly, that sort of addition is not possible - the addition operator is not applicable to string literals - so the rest of your question is invalid.

Asking a question "If X then why not Y?" when X is always false is pointless.

However, to clarify the error message from your compiler ....

Originally Posted by joaquim

why i can't:

Code:

string &operator+(const char *s1, const char *s2)
{
    string a=(string)s1+s2;
    return a;
}

???
heres the error message:
"'std::string operator+(const char*, const char*)' must have an argument of class or enumerated type"

The error message is self explanatory. At least one of the operands (i.e. the things being added if we're talking about addition) for any overloaded operator must be a user-defined type. Overloading operators for basic types (int, char, pointers to them, etc) is not permitted.

I've never seen the reason formally explained, but I assume it is to prevent programmers doing dumb things like overloading operators to make an expression like "2 + 3" give a result of 6701 (i.e. to make addition of basic types like int misbehave).

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