pass it array to function and back

#include <iostream>

using namespace std;

int parse ( int array );

int main ()
{
      
      int array[5];
      array[0] = 4;
      array[1] = 3;
      array[2] = 2;      
      array[3] = 1;
      array[4] = 0;
      
      array = parse ( array );
      
      cout << array[0] << endl;
      return 0;
}

int parse ( int array )
{
      
      array[0] = 0;
      array[1] = 1;
      array[2] = 2;      
      array[3] = 3;
      array[4] = 4;

      
      
      
      return array;
}




this doesn't work, how do I fix it?
TroudeloupAsked:
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Jaime OlivaresConnect With a Mentor Software ArchitectCommented:
in that case you return a copy of the original array, this could be helpful in some situations like:

instead of:
parse (array)
func1 (array)

you can do:
func1 (parse(array));

but I think it is not the case here.
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Jaime OlivaresSoftware ArchitectCommented:
you are not "receiving"  an array, it should be:

int *parse ( int *array )
{
     
      array[0] = 0;
      array[1] = 1;
      array[2] = 2;      
      array[3] = 3;
      array[4] = 4;
     
      return array;
}

but there are other problems:

array = parse ( array );
     
this is useless and not allowed, should be:

parse ( array );
     

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TroudeloupAuthor Commented:
yes i get it, because the function works on a pointer.

but i don't get why the function also has *

what is the * of ?




is this the fixed version of that?



#include <iostream>

using namespace std;

int parse ( int array );

int main ()
{
     
      int array[5];
      array[0] = 4;
      array[1] = 3;
      array[2] = 2;      
      array[3] = 1;
      array[4] = 0;
     
      parse ( array );
     
      cout << array[0] << endl;
      return 0;
}

int *parse ( int *array )
{
     
      array[0] = 0;
      array[1] = 1;
      array[2] = 2;      
      array[3] = 3;
      array[4] = 4;

     
     
     
      return array;
}


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Jaime OlivaresSoftware ArchitectCommented:
indeed it could be:

void parse ( int *array )
{
// etcetera
}

also, the prototype at the beginning always must match the real function:

void parse ( int *array );

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TroudeloupAuthor Commented:
what would it mean if the function has * ?
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Jaime OlivaresSoftware ArchitectCommented:
which one?
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TroudeloupAuthor Commented:
int *parse ( int *array )
{
     
      array[0] = 0;
      array[1] = 1;
      array[2] = 2;      
      array[3] = 3;
      array[4] = 4;
     
      return array;
}
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TroudeloupAuthor Commented:
not a single clue of what that does ;p
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