Double pointer as an array to a function [duplicate]

Question

This is a sample code of my program.

I have a function named function which is used to return an integer value and an integer array. I have to pass them as pointers (The signature of function cannot be changed). So I wrote the below code. Here I have to assign the values to the double pointer passed to the function.

void function(unsigned int* count, unsigned int ** array)
{
  for(unsigned int i = 0; i<4;i++)
  {
    //array[i] = (unsigned int*)i*10;
  }
  *count = 4;
}

int main()
{

  unsigned int count;
  unsigned int array2[4];
  function(&count, (unsigned int**)&array2);

  return 0;
}

But the above code is not working.

your program doesn't produce any observable output. how could it demonstrate your problem? at some point, you have to test it to see if it works. — Karoly Horvath
– Karoly Horvath, Commented Feb 12, 2014 at 10:38
My primary need was to fill the array 'array2' in the function 'function' with the specified values. I was debugging the code in VS to check whether it was working properly or not. — Aneesh Narayanan
– Aneesh Narayanan, Commented Feb 12, 2014 at 11:53

Marco A. · Accepted Answer · 2014-02-12 10:47:01Z

2

By reusing concepts I suppose you already know, you might use a pointer to the first element of the array and then pass the pointer's address to the function.

In that way you would have a double pointer to an unsigned int which is the first element of the array:

void function(unsigned int* count, unsigned int ** array)
{
  for(unsigned int i = 0; i<4;i++)
  {
    (*array)[i] = i*10; // WATCH OUT for the precedence! [] has higher precedence!
  }
  *count = 4;
}

int main(int argc, char* argv[])
{

  unsigned int count;
  unsigned int array2[4];
  unsigned int *pointer_to_array = &array2[0];
  function(&count, (unsigned int**)&pointer_to_array);

  return 0;
 }

As a sidenote:

If you could change the signature this would have made more sense:

void function(unsigned int* count, unsigned int * array)
{
  for(unsigned int i = 0; i<4;i++)
  {
    array[i] = i*10;
  }
  *count = 4;
}

int main(int argc, char* argv[])
{

  unsigned int count;
  unsigned int array2[4];
  function(&count, array2);

  return 0;
 }

The reason the above works is because when you pass an array into a function (directly or with an explicit pointer to that array) it essentially becomes a pointer. This is called array decaying.

edited Feb 12, 2014 at 10:47

answered Feb 12, 2014 at 10:38

Marco A.

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3 Comments

Paul R Over a year ago

He's not allowed to change the function signature apparently (see question for details) - probably homework?

Marco A. Over a year ago

Thanks Paul, I missed that one!

Aneesh Narayanan Over a year ago

@ Paul: Your comment seems funny, Anyway thanks for the support

BLUEPIXY · Accepted Answer · 2014-02-12 10:47:21Z

1

void function(unsigned int* count, unsigned int ** array){
    for(unsigned int i = 0; i<4;i++){
        (*array)[i] = i*10;
    }
    *count = 4;
}

int main(){
    unsigned int count;
    unsigned int array2[4];
    unsigned int *p = &array2[0];
    function(&count, &p);
    return 0;
}

answered Feb 12, 2014 at 10:47

BLUEPIXY

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Comments

Thomas Ruiz · Accepted Answer · 2014-02-12 10:38:41Z

0

To assign to an array that is passed by pointer, first dereference it, then assign the value.

void function(unsigned int* count, unsigned int ** array)
{
  for(unsigned int i = 0; i<4;i++)
  {
    (*array)[i] = i*10;
  }
  *count = 4;
}

answered Feb 12, 2014 at 10:38

Thomas Ruiz

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