Can I modify the target of a pointer passed as parameter?

问题

Can a function change the target of a pointer passed as parameter so that the effect remains outside the function?

void load(type *parameter)
{
    delete parameter;
    parameter = new type("second");
}

type *pointer = new type("first");
load(pointer);

In this minimal example, will pointer point to the second allocate object? If not, how can I get this kind of behavior?

Update: To clarify my intention, here is the code I would use if the parameter would be a normal type instead of a pointer. In this case I would simply use references.

void load(type &parameter)
{
    parameter = type("second");
}

type variable("first");
load(&variable);

That's easy but I try to do the same thing with pointers.

回答1:

No.

parameter will get a copy of the value of pointer in this case. So it is a new variable. Any change you make to it is only visible with in the function scope. pointer stays unmodified.

You have to pass the pointer by reference

void load(type *& parameter)
                ^
{

回答2:

You need to pass the pointer by reference:

void load(type *&parameter);

See for example http://www.cprogramming.com/tutorial/references.html

回答3:

Alternatively, you can use double pointers.

void load(type** parameter)
{
    delete *parameter;
    *parameter = new type("second");
}

type *pointer = new type("first");
load(&pointer);

But since you are using cpp you can use references

来源：https://stackoverflow.com/questions/16276176/can-i-modify-the-target-of-a-pointer-passed-as-parameter