initialize own String class by assignment in c++

Question

I'm writing my own string class StringEx in c++ (don't worry, just for exercise) but I'm failing at creating an instance of my class by assigning a string to it:

StringEx string1 = StringEx("test"); // works fine
StringEx string2 = "test"; // doesn't work

string string3 = "test";
string1 = string3; // also works fine

I overloaded the assignment operator so it can handle with std::string but I have to create an object of StringEx first.

How can I create a new object of StringEx by assigning a string to it? Is it even possible to get c++ handling every "string" as an object of my StringEx class?

This is my StringEx.h which works now

#ifndef STRINGEX_H
#define STRINGEX_H


#include <iostream>
#include <string>
#include <vector>
using namespace std; //simplyfying for now

class StringEx
{
private:
    vector<char> text;
public:
    StringEx();
    StringEx(string);
    StringEx(const char*);  // had to add this
    StringEx(vector<char>);

    int size() const;
    int length() const;
    char at(int) const;

    void append(const string&);
    void append(const StringEx&);
    void append(const char*);  // had to add this

    StringEx operator+(const string&);
    StringEx operator+(const StringEx&);
    StringEx operator+(const char*);  // had to add this too

    StringEx operator=(const string&);
    StringEx operator=(const StringEx&);
    StringEx operator=(const char*);  // had to add this too

    StringEx operator+=(const string&);
    StringEx operator+=(const StringEx&);
    StringEx operator+=(const char*);  // had to add this too

    friend ostream& operator<<(ostream&, const StringEx&);
};

#endif // STRINGEX_H

This is called initialization (more precisely, copy-initialization), not assignment. Assignment is when the object was already previously constructed (like in your last example). — interjay
– interjay, Commented Sep 2, 2015 at 8:57
You've already got some answers, but you really should show the declaration of your class's constructors, copy constructors and copy assignments. — Fabio says Reinstate Monica
– Fabio says Reinstate Monica, Commented Sep 2, 2015 at 9:15
@SimonKraemer MSVC has never been a paragon of standards compliance. — TartanLlama
– TartanLlama, Commented Sep 2, 2015 at 9:18
@LPrulzcrossover Your edit shows the StringEx.h that caused the error, or the StringEx.h that works now? Please edit your question and explain what you are showing. — Simon Kraemer
– Simon Kraemer, Commented Sep 2, 2015 at 9:24

JBL · Accepted Answer · 2015-09-02 09:07:42Z

7

A few precisions:

StringEx string1 = StringEx("test"); // works fine

This use the copy-constructor, i.e. StringEx(const StringEx& other);

StringEx string2 = "test"; // doesn't work

This tries to use a constructor with the following signature: StringEx(const char* str);

Finally, those two lines:

string string3 = "test";
string1 = string3; // also works fine

create an std::string from a const char*, which the standard library defines, and then use the copy-assignment operator overloaded for std::string that you seem to have defined correctly, i.e. StringEx& operator=(const std::string& other).

The key point here is that this statement:

Type myVar = something;

is not an assignement, it's an declaration and initialisation of a variable which uses a constructor, not the copy-assignment operator.

In your case, you're just missing a constructor that takes a const char* as parameter.

answered Sep 2, 2015 at 9:07

JBL

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5 Comments

Simon Kraemer Over a year ago

The original code can be compiled with MSVC14/VS2015 and works just fine. Even though the IDE shows an error.

SimonH Over a year ago

Thanks, works now! Is it possible to get c++ handling every const char* as a StringEx? In order to get i.e. cout << "hello" + "world"; work.

JBL Over a year ago

@LPrulzcrossover You can't implicitly (well, maybe you could in a very ugly way that I'm not aware of). Why would you want that anyway? That line can work with a proper overload of operator+ for two const char* that returns an std::string for example.

SimonH Over a year ago

Is it even possible to overload operator+ for two const char*? Considering this it seems like the answer is no, or how would you do that?

JBL Over a year ago

Should've checked before, it's actually not possible. C++ Standard at §13.5.6 specifies that at least one of the parameters type must be a class, a reference to a class, an enumeration, or a reference to an enumeration, when overloading binary operators. My bad.

TartanLlama · Accepted Answer · 2015-09-02 09:05:33Z

I assume that you have a non-explicit constructor of the form

StringEx (const std::string&);

or similar.

String literals are not of type std::string. "test" is of type const char[5]. std::string has a non-explicit constructor accepting a const char*, so your two calls look like this (not taking copy elision into account):

//implicitly convert "test" to temporary std::string
//construct temporary StringEx with temporary std::string
//copy-construct StringEx with temporary
StringEx string1 = StringEx("test"); 

//no StringEx constructor taking const char*
//two conversions necessary to make "test" into a StringEx
//invalid
StringEx string2 = "test";

An easy fix for this is to add a constructor taking a const char* to StringEx:

StringEx (const char*);

Alternatively you could just use direct initialization:

//only one conversion needed
StringEx string2 ("test");

Simon Kraemer · Accepted Answer · 2015-09-02 09:01:15Z

TLDR: The compiler won't autoresolve if 2 "casts" are necessary. In this case const char* to std::string to StringEx

StringEx string1 = StringEx("test"); // works fine > This calls the constructor explicitly and therefore only one conversion must be done: const char* to std::string.

StringEx string2 = "test"; // doesn't work > This on the other hand is not clear. Do you want to cast const char* to std::string and use the StringEx(const std::string&)constructor or use some other intermediate class?

What you need is another constructor accepting const char* as parameter.

BTW:

StringEx string4 = std::string("test"); //Also works

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