C ++将指针传递到临时对象（堆上）到函数的方式？ [英] C++ Ways to pass a pointer to a temporary object (on the heap) to a function?

问题描述

我有一个函数，它接受一个指向自定义类的对象的指针（实际上是指向一个基类的指针，这样多态可以工作）。然而，在呼叫例程中，为了该呼叫的目的，该对象是唯一需要的，即是临时的。例如像这样：

I've got a function that takes a pointer to an object of a custom class (actually pointers to a base class such that polymorphism works). Within the calling routine this object however is exclusively needed for the purpose of this call, i.e. is temporary. For example like this:

class A { /** stuff */ };
class B : public A { /** stuff */ };

void doSomething( const A* const _p ) { /** stuff */ }

void callingRoutine()
{
  A* tempPointer = new B;
  doSomething( tempPointer );
  delete tempPointer;
}

现在，因为我真的只需要 doSomething 的调用中有> B ，有没有办法在一行？正在做

Now, since I really only need the object of type B within the call to doSomething, is there a way to do it in one line? Doing

doSomething( new B );

会创建内存泄漏（valgrind说）。或

creates memory leaks (valgrind says so). Or would

doSomething( &B );

后者编译但是给出关于将指针传递给临时对象的警告。这是我想做的，但是这样会安全吗？

be the recommended way? The latter compiles but gives warnings about passing pointers to temporary objects. This is what I want to do, but would it be safe this way?

推荐答案

>

The cleanest way is to do

B b;
doSomething(&b);

但是你真正应该写什么取决于 doSomething 函数。如果在 callRoutine 结尾处破坏 b 很好，那么这是更快更干净的方法，因为在堆栈上的分配速度快于 new ，并且不要求您之后删除 b 。

But what you really should write depends of what the doSomething function does. If it's fine to destruct b at the end of callingRoutine, then it's the faster and cleaner way to do this, because allocating on the stack is faster than new and does not require you to delete b afterwards.

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