Hi, I have a question about 8051
I have an instruction which at adress 0020H and codded as 80F0H
what would be the value of PC?
i don't understand how can i decode it

 

Hi, I have a question about 8051
I have an instruction which at adress 0020H and codded as 80F0H
what would be the value of PC?
i don't understand how can i decode it

You need to establish which of two possible orderings a 16-bit word can be placed in a byte addressable memory. It has been too long since I had a need to know this, and I cannot remember which way it went. Other members of the Intel family put the least significant byte of a word in the lowest address. After the instruction is fetched, the PC will be at address 00022H.

 

because you are within the interrupt vectors table, it really depends on the instruction at the entry point of the interrupt.

 

The interrupt vector table begins a 0000H.
Address 0000H is a 3-byte entry which normally would be a LJMP instruction.

Assuming that you are not using the rest of the table for interrupts, the instruction
80 F0 = SJMP F0

Since SJMP is a relative jump, it would be executed as SJMP (PC + 2 - 16)
Since PC = 0020H = 32
PC will become 0012H

 

Thank you so much all for the explanation !

 

The interrupt vector table begins a 0000H.
Address 0000H is a 3-byte entry which normally would be a LJMP instruction.

Assuming that you are not using the rest of the table for interrupts, the instruction
80 F0 = SJMP F0

Since SJMP is a relative jump, it would be executed as SJMP (PC + 2 - 16)
Since PC = 0020H = 32
PC will become 0012H

Can you please explain from where did you find 16?

 

It is not 16. It is -16.
The instruction opcode is 80 F0.
80 is the opcode for SJMP, meaning Short Jump.
F0 is the 8-bit relative address. This is encoded in an 8-bit signed integer having values from -128 to plus 127.
7F = 127
7E = 126
:
01 = 1
00 = 0
FF = -1
FE = -2
:
F0 = -16
:
81 = -127
80 = -128