Re: conversion from '...' to non-scalar type '...' requested

From:
"Jonathan Mcdougall" <jonathanmcdougall@gmail.com>
Newsgroups:
comp.lang.c++
Date:
5 Jun 2006 12:17:14 -0700
Message-ID:
<1149535034.339342.207280@y43g2000cwc.googlegroups.com>
tthunder@gmx.de wrote:

Hi @all,

My small example does not compile... I know, that this (as always) has
reasons, but I want to know WHY?


Because you cannot bound an rvalue to a non-const reference.

BTW:
I only get errors with g++ (4.x), BCB (6.0),...
VS C++ (2005) works perfectly (without warnings etc.)

class classValue
{
        public:
                classValue() {}
};

class classHolder
{
        public: classHolder(classValue &ami) {} // <== I know, but in
this case reference MUST be non-const
};

classHolder getAHolder()
{
        return classValue(); // <--- Error: conversion from
'classValue' to non-scalar type 'classHolder' requested


If you could do that, you'd get several more important problems:

void f(int& i)
{
  i = 2;
}

int main()
{
  int i = 4;
  double d = 1.0;

  f(i); // ok
  f(d); // ok??
  f(4); // ouch!

  // what should this output?
  std::cout << d;
}

For the second call to f(), the double gets converted to a temporary
int (an rvalue), which gets modified in f(). Do you expect 'd' to get
modified also? It won't and that's misleading, dangerous and illegal.

}

Using a little trick, everthing works well...


Yes, you may do that.

class classValue
{
        public:
                classValue() {}
                classValue &self() { return *this; }
};

class classHolder
{
        public: classHolder(classValue &ami) {}
};

classHolder getAHolder()
{ return classValue().self();}

But why must I use this nasty indirection?


Because that's how the language is defined. Note that you may also do

classholder getAHolder()
{
  classValue v;
  return classHolder(v);
}

Jonathan

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