MP Compares the first source operand with the second source operand and sets the status flags in the EFLAGS register according to the results. The comparison is performed by subtracting the second operand from the first operand and then setting the status flags in the same manner as the SUB instruction. When an immediate value is used as an operand, it is sign-extended to the length of the first operand. So, only the flags are affected. Operation: temp = Source1 - SignExtend(Source2); ModifyStatusFlags(); //Modify status flags in the same manner as the SUB instruction Flags affected: The CF, OF, SF, ZF, AF, and PF flags are set according to the result.
branch instruction
INR increment the content of register/memory by 1and result is stored in same place. INX increment the register pair by 1(no flags are affected)
The compare and subtract instructions in the 8085 both subtract one operand from another, and set flags accordingly. The subtract instruction stores the result in the accumulator, while the compare instruction does not - except for the flags, the compare instruction "throws" the result away.
To reset the pending RST 7.5 instruction in the 8085, you need to execute a SIM instruction with a particular value in the accumulator. PUSH FLAGS MVI A,10H SIM POP FLAGS Of course the PUSH and POP are optional, if you don't need to preserve the value of the accumulator.
To add a new machine language instruction to an processor instruction set, you need to replace the microcode of the processor.
All flags are affected after the SUB operation to reflect the result of subtraction.
The NOP instruction is short for no-operation. It is an executable instruction that does nothing to the processor, its registers, or its flags. It is useful in timing loops, or to provide room for patchabilty of a piece of code.
add B
the compare instruction of 808 is used to compare the 2 operands. syntax: cmp op1,op2 algorithm: op1-op2 the value of the operands are not affected only the flags are updated if op1<op2 carry=1 and zero flag=0 if op1=op2 cy=0 and zf=1 if op1>op2 cy=0 and zf=0
That would depend on the computer architecture.
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