在C++中,有一个函数void foo(int**p)的用法是什么？

Question

我的同事告诉我,void foo(int** p)它在C#中像out参数一样使用.有人可以准确解释如何？

我明白了,但是有些东西不见了.我知道如果我们将指针p本身传递给foo(*p)函数体p = new int(),我们可能会有一个悬空修改器!但是如何foo(**p)阻止这样的事情发生呢？

Answer 1

void foo(int** p)
{
 *p = new int;
}


void main()
{
 int *p = NULL;

foo(&p);
//unless you pass the pointer to p, the memory allocated in the 
//function foo is not available here.

foo(p); // lets assume foo is foo(int* p)

//when you pass only pointer p to `foo()` then the value of the pointer will be 
//passed ( pass by value) to foo() and hence, any modification to pointer p ( in
// the form of allocating memory also) will not be available after the
// control returns from foo()
// The p will still points to NULL here, not the memory allocated in foo().  

}