I modified a bit your code to guarantee the operation ;

test1 =3D 512;

writeEepromByte(0x00, test1t&0xFF);
Delay10TCYx(200);
writeEepromByte(0x01, test1>>8);
Delay10TCYx(200);
test2 =3D readEepromByte(0x00+ (readEepromByte(0x01)<<8));
Delay10TCYx(100);
test2 =3D test2 + (readEepromByte(0x01)<<8);
Delay10TCYx(100);

f( test1=3D=3Dtest2) {
 PORTBbits.RB7 =3D1; }

Now RB7 is low always, it's not changing into high state.

On Fri, Oct 28, 2011 at 5:26 PM, Kerry Wentworth
<kwentworth@skunkworksnh.com> wrote:
> You can write/read a 16 bit integer (512 requires a 16 bit integer) by:
>
> // Write
> test1 =3D 512;
> writeEepromByte(0x00, test1&0xFF);
> writeEepromByte(0x01, test1>>8);
>
> // Read
> test2 =3D readEepromByte(0x00)+(readEepromByte(0x01)<<8);
>
> // Check
> if( test1=3D=3Dtest2) {
> PORTBbits.RB7 =3D1; }
>
> Kerry
>
>
> Yigit Turgut wrote:
>> Hi all,
>>
>> I am trying to write an 8bit integer to internal eeprom of a 18f2550
>> by the following functions ;
>>
>> #define byte unsigned char
>> #define BYTE_ADDR 0
>> int test1, test2;
>>
>> void writeEepromByte(byte addr, byte data) {
>> INTCONbits.GIE =3D 0;
>> EECON1bits.EEPGD =3D 0;
>> EECON1bits.CFGS =3D 0;
>> EECON1bits.WREN =3D 1;
>> EEADR =3D addr;
>> EEDATA =3D data;
>> EECON2 =3D 0x55;
>> EECON2 =3D 0xaa;
>> EECON1bits.WR =3D 1;
>> while (!PIR2bits.EEIF) ;
>> PIR2bits.EEIF =3D 0;
>> INTCONbits.GIE =3D 1;
>> }
>>
>> byte readEepromByte(unsigned char addr) {
>> EECON1bits.EEPGD =3D 0;
>> EECON1bits.CFGS =3D 0;
>> EEADR =3D addr;
>> EECON1bits.RD =3D 1;
>> return EEDATA;
>> }
>>
>> And to test it I use ;
>>
>> if ( PORTBbits.RB2 =3D=3D 0 ){
>> Delay10KTCYx( 50 );
>> test1 =3D 512;
>> writeEepromByte(0x00, test1);
>> Delay10TCYx(200);
>> test2 =3D readEepromByte(0x00);
>> Delay10TCYx(100);
>> }
>>
>> if( test1=3D=3Dtest2<<8) {
>> PORTBbits.RB7 =3D1; }
>>
>> All ports are configured correctly and board is functioning well. With
>> a push of the button at RB2, value 512 should be stored in eeprom.
>> Actually least significant 8bits of test1 will be written because it's
>> a one-bye space that it's writing to. Even if so, =A0test1=3D=3Dtest2<<8
>> this expression should be valid and RB7 is should be set high. None of
>> this happens, nothing happens at all. Is this a problem because only
>> bytes can be written to eeprom ? It should be pretty straightforward
>> to write/read 8 bit integer to/from internal eeprom/
>>
>
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