A Question On Reference Type

**laserlight** · 08-14-2008

No. The former declares a function named t that takes no arguments and returns a Test. The latter is not allowed, as noted by MarkZWEERS.

**siavoshkc** · 08-15-2008

How the constructor is called without using 'new'? Is it implicity called when you call the constructor? It seems by calling the constructor weare implicity using 'new', am I right?

The latter is not allowed, as noted by MarkZWEERS.

Would you explain more, please?

**laserlight** · 08-15-2008

How the constructor is called without using 'new'? Is it implicity called when you call the constructor?

Whenever you create an object, one of the object's constructors is implicitly called. Even the use of new does not explicitly call a constructor (and I believe that applies even to placement new).

It seems by calling the constructor weare implicity using 'new', am I right?

No.

Would you explain more, please?

Attempt to compile this program:

Code:

class Test {};

int main()
{
    Test &t = Test();
}

**siavoshkc** · 08-15-2008

This code compiled and ran successfully:

Code:

#include <iostream>

using namespace std;


class Test {
public:
    Test() { cout << "Test" << (int)this << endl; }
    ~Test() { cout << "~Test" << endl; }
      void test(){cout <<"fun"<<endl;}
};

int main()
{
    Test& t = Test();
      t.test();
    cout << "main" << endl;

    return 0;
}

[edit]

Whenever you create an object

Where did we create an object in

Code:

Test();

?

**laserlight** · 08-15-2008

This code compiled and ran successfully:

Disable compiler extensions. Your code should not compile on a standard compliant implementation.

**siavoshkc** · 08-15-2008

Disable compiler extensions.

Yes now it is giving me two errors

delme.cpp(8) : error C2440: 'type cast' : cannot convert from 'Test *const ' to 'int'
The target is not large enough
delme.cpp(15) : error C2440: 'initializing' : cannot convert from 'Test' to 'Test &'
A non-const reference may only be bound to an lvalue

What is this extension?

**Elysia** · 08-15-2008

Visual C++ is a known compiler to allow this extension. If you go to warnings level 4, you should also receive a warning saying non-standard extension used.
The extension allows you to bind a non-const reference to a temporary, which is disallowed by the standard (only const references can bind to temporaries, which has been pointed out earlier).

**siavoshkc** · 08-15-2008

OK. When we write

Code:

Test()

What happens? Does a memory allocation take place for class on heap or it is just a temp on stack (or somewhere else)? Why it can be bind to const ref but not non-const?

**Elysia** · 08-15-2008

Unless you use new, it's on the stack. Always the stack, unless it's new.
I believe it's because you shouldn't be able to modify a temporary because it's just that - a temporary. Its lifetime is less than that of the variable it's bound to. Bad things would happen if you try to modify it after its lifetime has expired (while the lifetime of the variable holding the reference is still alive).

**siavoshkc** · 08-15-2008

When I disabled extension it generates error for my type cast. So how should I cast that?
Second this code works fine:

Code:

class Test {
public:
    Test() { cout << "Test"  << endl; }
    ~Test() { cout << "~Test" << endl; }
      void test(){cout <<"fun"<<endl;}
};

int main()
{
      Test();
    Test t = Test();
      t.test();
    cout << "main" << endl;

    return 0;
}

Why by calling constructor it takes place in stack?

[EDIT]And what about this

Code:

class Test {
public:
    Test() { cout << "Test"  << endl; }
    ~Test() { cout << "~Test" << endl; }
      void test(){cout <<"fun"<<endl;}
}rt;

rt is static in stack in file scope. Right?

**cpjust** · 08-15-2008

That's one thing that has burned me a number of times... What exactly is the lifetime of a temporary object? It's obviously less than a regular variable which ends when it gets to its closing brace.
I think this is one of the cases where I've gotten burned:

Code:

std::string Func() { ... }

int main()
{
   ...
   AnotherFunc( Func().c_str() );
   ...
}

In the case above, I believe the temporary object doesn't even life until the end of the function call.

**laserlight** · 08-15-2008

When I disabled extension it generates error for my type cast. So how should I cast that?

What type cast?

Why by calling constructor it takes place in stack?

What is the stack?

**cpjust** · 08-15-2008

Originally Posted by siavoshkc

Code:

class Test {
public:
    Test() { cout << "Test"  << endl; }
    ~Test() { cout << "~Test" << endl; }
      void test(){cout <<"fun"<<endl;}
};

int main()
{
      Test();
    Test t = Test();
      t.test();
    cout << "main" << endl;

    return 0;
}

Why by calling constructor it takes place in stack?

That line (in green) is completely useless since you're calling the default constructor twice. Just do this:

Code:

Test t;

**Daved** · 08-15-2008

>> When I disabled extension it generates error for my type cast. So how should I cast that?
Why do you need a cast there? Just output this (or cast to void*).

Temporaries passed to a function live until the function ends, but in cpjust's example the result of Func() isn't passed to a function, a pointer is. So the result of Func() lives until c_str() is called and it's value is sent to the function. Then it can be destroyed.

**CornedBee** · 08-15-2008

A temporary lives until the end of the "full-expression", i.e. the next semicolon. Unless, that is, it is bound to a reference, in which case its lifetime is extended to that of the reference.

In particular, a temporary appearing with function parameters lives until the full expression containing the function call finishes. However, if the function keeps a reference to the object in question, this reference will soon be invalid.