writeEepromByte(0x01, result>>8);

this line is ;

writeEepromByte(0x01, test1>>8);

I mistyped.

On Fri, Oct 28, 2011 at 7:01 PM, Yigit Turgut <y.turgut@gmail.com> wrote:
> test1 =3D 512;
>
> if ( PORTBbits.RB2 =3D=3D 0 ){
> Delay10KTCYx( 50 );
> writeEepromByte(0x00, test1&0xFF);
> Delay10TCYx(200);
> writeEepromByte(0x01, result>>8);
> Delay10TCYx(200);
> test2 =3D readEepromByte(0x00 + (readEepromByte(0x01)<<8));
> Delay10TCYx(100);
> }
>
> if( if test1 !=3D test2 ){
> LATBbits.LATB7 =3D 1;
> }
>
> RB7 is always on now where it shouldn't be.
>
> On Fri, Oct 28, 2011 at 6:40 PM, Kerry Wentworth
> <kwentworth@skunkworksnh.com> wrote:
>> You are reading the high byte twice, and adding them together. =A0Elimin=
ate:
>> test2 =3D test2 + (readEepromByte(0x01)<<8);
>> Delay10TCYx(100);
>>
>> Kerry
>>
>>
>>
>> Yigit Turgut wrote:
>>> 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) {
>>> =A0PORTBbits.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. Wit=
h
>>>>> 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 o=
f
>>>>> 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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>>>> View/change your membership options at
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>>>>
>>>>
>>>
>>>
>>
>> --
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>>
>

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