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Returning pointer to int32

 
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Guest
Returning pointer to int32
PostPosted: Sat Jan 31, 2009 10:51 am     Reply with quote
What is wrong here?

It's only a small test program to show my problem.
I want to return a pointer to a int32.
I have another function returning a string it work.

Cant compile.

Code:
int32 serialno;

int8 *serial_no(){
 return (serialno);
}

printf("%Lu",serial_no);



working:

Code:
char serialstr[10];

int8 *serial_str(){
 return (serialstr);
}


printf("%s",serialstr);
Guest
PostPosted: Sat Jan 31, 2009 10:54 am     Reply with quote
Sorry right type.
printf("%Lu",serial_no());
printf("%s",serial_str());
Ttelmah
Guest
PostPosted: Sat Jan 31, 2009 2:04 pm     Reply with quote
Code:

int32 serialno;

//first the function is returning a pointer to an int32, not an int8....
int32 *serial_no(){
  //second, the pointer, is the address of the variable, not the variable
  //itself.
  return (&serialno);
}

//Third, the print, needs to reference the contents of the pointer
printf("%Lu",*serial_no());

On your 'does compile' version, you are getting away with using the array name, since in C, _when dealing with an array_, the name of the array is a shorthand, for it's address.
Your print then accesses the array, not the function though.....

Best Wishes
Guest
PostPosted: Sun Feb 01, 2009 4:15 am     Reply with quote
Thanks

But something I dont understand is what type to select in the pointer int, int16, int32...

Ex.
Why not working with int8? When the string function is working.

Code:
int32 serialno;

[b]int8[/b] *serial_no(){
 return (&serialno);
}

printf("%Lu",*serial_no());




This is working, but is it right.
I don't use "&" because the str[] is a pointer right?

Code:
char str[11];

[b]int8 [/b]*serial_str(){
 return (str);
}

printf(serial_str()); //must I use "*"? It work without.


Confused:-(
FvM



Joined: 27 Aug 2008
Posts: 2337
Location: Germany

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PostPosted: Sun Feb 01, 2009 10:06 am     Reply with quote
The compiler tells why: It is expecting an int32 expression, but *serial_no() is int8. You didn't tell, why do you want use a wrong pointer type?

If meaningful at all, you can do with a type cast:
Code:
printf("%Lu",*(int32*)serial_no());
Ttelmah
Guest
PostPosted: Sun Feb 01, 2009 12:33 pm     Reply with quote
Key to understand, is that a 'char', _is_ an int8 (in CCS). Int, int8, char, and byte, are all exactly the same type. Hence the 'int8' pointer, works as expected.
The name of an array, is a 'C shortcut', to the pointer to it, so in your example, 'str', is equivalent to '&str[0]'. However this only works with arrays.

Best Wishes
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