CODEWITHSUNDEEP
c++objectconstructorstack
Mister Smith
Mister Smith

Reputation: 28168

How to reuse object created in the stack

I'm not a C++ programmer, and I have great respect for its complexity, so I'd normally stay away from it. Despite this, I've found a certain library I'd like to tinker with.

The library includes a sample snippet where a single instance is allocated in local scope:

    Foo foo(arg1, arg2, ...);

I want to modify this example to re-initialize the object and do more things with it, but I can't do it via getters/setters or public variables. Is this possible without creating more instances?

In case there's no way of doing that without creating new instances, how could I release the memory of the objects I no longer need?

For instance, if it were a dynamic object, I guess it would be something like this:

    Foo* pfoo;

    pfoo = new Foo(arg1, arg2, ...);
    pfoo->doSomething();
    delete pfoo;

    pfoo = new Foo(arg3, arg4, ...);
    pfoo->doSomething();
    delete pfoo;

What would be the equivalent for the object allocated in the stack?

Upvotes: 2

Views: 2003

Answers (3)

Richard Hodges
Richard Hodges

Reputation: 69884

Just wanted to post one more alternative for the OP, more along the lines of what he has suggested:

void somefunction()
{
    // create the first foo owned by a smart pointer
    std::unique_ptr<Foo> pFoo { new Foo { arg1, arg2, ... } };
    pFoo->doSomething();

    // now replace the old Foo with a new one
    pFoo.reset( new Foo { arg3, arg3, ... } );
    pFoo->doSomething();

    // no need to delete anything unless you specifically want the Foo to die now
    // in which case...
    // pFoo.reset();
} // implicit destruction of whatever pFoo owns happens here

Upvotes: 2

nevermind
nevermind

Reputation: 2438

If an object doesn't allow to reinitialize itself via public setters - you can not do it.

Actually, in your example you are not reinitializing object, you create another instance of Foo and reinitialize a pointer variable with address of this object.

Lifetime of local variable is limited to their scope. So you can split the scope of your function into smaller scopes:

void bar()
{
  {
    Foo foo(arg1, arg2, ...);
    foo.doSomething();
  }
  {
    Foo foo(arg3, arg4, ...);
    foo.doSomething();
  }
}

Upvotes: 2

Jeff
Jeff

Reputation: 3525

The only way to release automatic variables is to let them fall out of scope:

void bar() {
  Foo f1(arg1, arg2);
  f1.doSomething();

  {
    Foo f2(arg3, arg4);
    f2.doSomething();
    // other stuff ...
    // f2 dies here.
  }

  {
    Foo f3(arg5, arg6); // allocated on stack, possibly overlapping f2's old spot
    f3.doSomething();
    // ...
    // f3 dies here.
  }

  // f1 dies here.
}

Upvotes: 9

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