Problem transferring an array to a routine

[Revo]

When I compile with PCM and PIC16f876 it work fine

But when i compile with PCH v3.168 and PIC18F252 the array is not transferred correctly
number[1] SHOULD BE 2 BUT IT IS ANOTHER VALUE




void routine(int number[5])
{
printf(lcd_putc," %u",number[1]);//<-------number[1] SHOULD BE 2 BUT IT IS ANOTHER VALUE
}

main (){
int array[5];
lcd_init();

array[1]=2;
printf(lcd_putc," %u",array[1]);
routine(array);

}


Can anyone help me?

[PCM programmer]

I modified your program slightly, because I don't have a LCD on
my 18F458 demo board. I compiled it with PCH vs. 3.178, and
it worked OK. It displayed this on the terminal window:

2 2

I don't have PCH vs. 3.168, so I can't test that version.

[revo]

I have run your code in my demo board

It displayed this on the terminal window 2 0

So I think it is a bug in the version 3.168

I will update my compiler

Thanks

[Revo]

I have updated my CCS
But I have the same problem. The terminal displayed 2 0

I found a solution to resolve it.
The follow code work Ok


#include <18F458.h>
#device ICD=TRUE
#fuses HS, NOWDT, NOPROTECT, BROWNOUT, PUT, NOLVP
#use delay(clock = 4000000)
#use rs232(baud = 9600, xmit=PIN_C6, rcv = PIN_C7, ERRORS)
//========================================
void routine(int number)// <---- This is the difference before: void routine(int number[5])
{
printf(" %u",number[1]);
}

main ()
{
int array[5];

array[1]=2;
printf(" %u",array[1]);
routine(array);

while(1); // Prevent PIC from going to Sleep
}

[Gerrit]

Revo,

Should it not be an pointer that you receive in your fuction?


void routine(char * number)//
{
printf(" %u",number[1]);
}



Gerrit

[PCM programmer]

That is strange. I have two different demo boards.
On the one I tested with your code earlier, I get "2 2".
On my other demo board, I get "2 0", just like you did.
Something weird is going on. In each case, I am using
PCH vs. 3.178. I don't have time to research it right now,
but I will, and I'll post the reason if I can find one.