Passing a pointer to a pointer? Topic is solved

[spamiam]

I want to pass a pointer to an array to a newly instantiated class. This new class will determine how big the array needs to be, allocates it with "new" and fills it. Then the original class needs to know where that array is located.

I think I need to pass a pointer to the pointer, but I do not know how to do this properly.

-Tony

[Auria]

Something like (disclaimer: untested, used only to give an idea) :

Code: Select all

void foo()
{
    int* myarray;
    bar( &myarray );
}


void bar(int** array_ptr)
{
    *array_ptr = new int[128];
}

[spamiam]

void bar(int** array_ptr)
{
*array_ptr = new int[128];
}

Auria,

Thanks. I just did it that way and it works! What I had hoped to be able to do was use a more simple name for that double-indirection. I wanted to be able to refer to the pointer "*array_ptr" as something like "array".

It is harder and less intuitive to write

Code: Select all

*array_ptr[1] = 23;

than it is to write

Code: Select all

array[1] = 23;

I realize that I could do this with a macro, but I find macros hard to maintain, and I bet that this would most definitely be one of those cases. Any suggestions?

[Auria]

You could simply do something like :

Code: Select all

void foo()
{
    int[] myarray;
    bar( myarray );
}


void bar(int[]& array_ref)
{
    array_ref = new int[128];
    array_ref[0] = 123;
    array_ref[1] = 345;
    // etc...
}

(again, same disclaimer, untested)
Or again :

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void foo()
{
    int* myarray = bar();
}


int* bar()
{
    int* array_ptr = new int[128];
    array_ptr[0] = 123;
    array_ptr[1] = 345;
    return array_ptr;
}

[spamiam]

You could simply do something like :

Code: Select all

void foo()
{
    int[] myarray;
    bar( myarray );
}


void bar(int[]& array_ref)
{
    array_ref = new int[128];
    array_ref[0] = 123;
    array_ref[1] = 345;
    // etc...
}

Auria,

THAT is what I was looking for! I had come to the conclusion by checking around that I needed to use the ampersand in the args of the constructor, but I was not sure how to go about doing it.

I was surprised by the brackets in the arg. I had been thinking that maybe I needed to code it something like this.

Code: Select all

void foo()
{
    int *myarray;
    bar( &myarray );
}


void bar(int* &array_ref)
{
    array_ref = new int[128];
    array_ref[0] = 123;
    array_ref[1] = 345;
    // etc...
}

I have not tried this exact version, would this work too?

I have never seen a definition written as "int[] varname" (I am familiar with int *varname), but my experience in C++ is fairly limited, especially with more advanced pointer and reference topics.

-Tony

[Auria]

In C++, int[] and int* are often used as synonyms (int* is not necessarily an array, but pointer semantics allow to call [n] on a pointer so the effect is essentially the same)