Strange assignment problem - C++

Question

Having a strange problem making a new string class and assigning an array of char* to it in the GCC compiler. Source code:

#include "../Include/StdString.h"

StdString::StdString()
{
    //ctor
    internstr = std::string();
}

char* StdString::operator=(StdString other) {
    return other.cstr();
}

StdString StdString::operator+(StdString other) {
    StdString newstr = StdString();
    newstr.internstr = internstr+other.internstr;
    return newstr;
}

void StdString::operator=(char* other) {
    internstr = other;
}

StdString::~StdString()
{
    //dtor
}

char* StdString::cstr() {
    return (char*)internstr.c_str();
}

Error: conversion from char* to non-scalar type StdString requested.

How did std::string do their assignments?

Code wasn't pasting right from my IDE like it usually would (indents were incorrect), so I thought it would be quicker to use pastebin — bbosak, Sep 21 '11 at 02:30
Did you get the error on this method: void StdString::operator=(char* other) ? I feel strange on this operator overloading. — dip, Sep 21 '11 at 02:31

score 2 · Accepted Answer · edited May 23 '17 at 12:03

2

std::string can do its conversion because it defines a conversion constructor. Something like this.

class std::string {
  // ...
std::string(const char *);
};

NOTE: the actual std::string is more complicated.

With the assignment operator you should be able to do

StdString str;
str = "hello";

but not

StdString str = "hello";

edited May 23 '17 at 12:03

Community

1
1

answered Sep 21 '11 at 02:30

David Nehme

21,379
8
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If `StdString` is more like `std::string`, then the initialization must also work. `std::string` has a overloaded constructor having argument type `const char*`. [Reference](http://www.cplusplus.com/reference/string/string/string/) – Mahesh Sep 21 '11 at 02:40
Thanks! This turned out to be the solution – bbosak Sep 21 '11 at 02:46

wilhelmtell · Answer 2 · 2011-09-21T02:51:27.170

The error you're having is for not providing a constructor that takes a char*. This is the conversion function the compiler complains about for being missing.

StdString::StdString(char const* s)
{
    // ...
}

Also, if your internal string is a std::string then you don't need any of the assignment operators, copy constructor and destructor. Once you add the char* conversion constructor the compiler-provided assignment operator will magically work for char* as well. Well, not really magically: the compiler will see that it can convert a char * to a StdString through your conversion constructor, and then use that with the implicit assignment operator.

You can also leave the default constructor's definition empty; this will give you the default construction for all members, which is probably good enough.

score 1 · Answer 3 · answered Sep 21 '11 at 02:46

1

StdString mystr = "Hello world!";

requires copy constructor.

Try to add the following:

StdString::StdString(const char* other)
{
    internstr = other;
}

answered Sep 21 '11 at 02:46

dip

530
3
7

score 0 · Answer 4 · answered Sep 21 '11 at 02:37

0

Looks like you are confusing assignment and initialization. Initialization uses a constructor, even when it's called with a "=" symbol.

answered Sep 21 '11 at 02:37

Adam Mitz

6,025
1
29
28

score 0 · Answer 5 · answered Sep 21 '11 at 02:42

Why do you define your conversion operator as:

char* StdString::operator=(StdString other) 
{ 
    return other.cstr(); 
}

All this does return the contents of other without setting the current class's internstr to that given by other.

What I would do would look more like:

StdString& StdString::operator=(StdString other) 
{ 
    //  copy contents of other to this->internstr

    return *this;
}

Strange assignment problem - C++

5 Answers5