Solved

pass it array to function and back

Posted on 2007-11-30
8
182 Views
Last Modified: 2010-04-01
#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?
0
Comment
Question by:Troudeloup
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 4
  • 4
8 Comments
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20386745
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 );
     

0
 

Author Comment

by:Troudeloup
ID: 20386750
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;
}


0
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20386767
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 );

0
Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 

Author Comment

by:Troudeloup
ID: 20386776
what would it mean if the function has * ?
0
 
LVL 55

Expert Comment

by:Jaime Olivares
ID: 20386789
which one?
0
 

Author Comment

by:Troudeloup
ID: 20386811
int *parse ( int *array )
{
     
      array[0] = 0;
      array[1] = 1;
      array[2] = 2;      
      array[3] = 3;
      array[4] = 4;
     
      return array;
}
0
 
LVL 55

Accepted Solution

by:
Jaime Olivares earned 500 total points
ID: 20386826
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.
0
 

Author Comment

by:Troudeloup
ID: 20386829
not a single clue of what that does ;p
0

Featured Post

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

In days of old, returning something by value from a function in C++ was necessarily avoided because it would, invariably, involve one or even two copies of the object being created and potentially costly calls to a copy-constructor and destructor. A…
Templates For Beginners Or How To Encourage The Compiler To Work For You Introduction This tutorial is targeted at the reader who is, perhaps, familiar with the basics of C++ but would prefer a little slower introduction to the more ad…
The goal of the tutorial is to teach the user how to use functions in C++. The video will cover how to define functions, how to call functions and how to create functions prototypes. Microsoft Visual C++ 2010 Express will be used as a text editor an…
The viewer will learn how to user default arguments when defining functions. This method of defining functions will be contrasted with the non-default-argument of defining functions.

728 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question