uVision3 Bug?

why is the simple code below assembled wrong?

Because it isn't. There's nothing wrong with that code. Dan already told you why:

The disassembly copies the same value from the same memory location 0x06 to tp(n). (MOV @R0,0x06)

Memory location 0x06 is the location of (the bank 0 rendition of) the R6 register, so there is nothing wrong with this: it copies the content of R6 (which at this moment holds tp[n-1]) to where R0 currently points (tp[n])

So whatever the actual problem is, this code snippet is not it.

why is the simple code below assembled wrong?

Because it isn't. There's nothing wrong with that code. Dan already told you why:

The disassembly copies the same value from the same memory location 0x06 to tp(n). (MOV @R0,0x06)

Memory location 0x06 is the location of (the bank 0 rendition of) the R6 register, so there is nothing wrong with this: it copies the content of R6 (which at this moment holds tp[n-1]) to where R0 currently points (tp[n])

So whatever the actual problem is, this code snippet is not it.

Thanks for your answers, but i still have the problem.
I tried to shorten the code as much as possible.
If you replace the function call "get_taste()" with the code within the get_taste-function-code then it works ok. But i don't understand why. There should be no difference.
Christian Tauschek

unsigned char tp[2];
unsigned char n, r1;

void timer1_isr(void) interrupt 3 using 1 {
     TL1 = 0xef;
     TH1 = 0xd8;

     // if i replace this call "get_taste()" with
     // the get_taste-code below, then it works ok
     get_taste();
}

void get_taste(void) {
     tp[0x00] = 0x0a;
     tp[0x01] = 0x0b;
     n = 0x01;
     tp[n] = tp[n - 1]; // there is the MISTAKE

     // now i expect the value 0x0a in r1, but r1
     // has the value 0x00
     r1 = tp[n];
}

Note the use of "using 1" which selects a specific register bank.

But how would get_taste() know that you call it from an ISR that is "using 1"?

When you mix code using different register banks, then you can get these effects.

Memory location 6 is always at the same place, but the definition of R6 will change depending on the currently used register set (the "using" part).

void get_taste(void) {
     tp[0x00] = 0x0a;
     tp[0x01] = 0x0b;
     n = 0x01;
     tp[n] = tp[n - 1]; // there is the MISTAKE

No. That is only where the mistake caused elsewhere finally bites.

There is a pair of places where the mistake is: you call a function from an ISR qualified "using 1", but the called function itself is not qualified "using 1". That combination is a monstrously big NO-NO.

Hello all,
thanks for answers. Now, without "using 1", it works fine.
I caused the mistake.

Christian Tauschek

However, using significantly reduces the overhead in your ISR - so you should really know how to use it properly (even if it isn't needed in this particular case)

See http://www.keil.com/support/man/docs/c51/c51_le_regbankaccess.htm for details.

Also: http://www.keil.com/support/man/docs/c51/c51_le_regbankspec.htm and the linked knowledgebase articles.