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CMP: Compare: - This instruction compares the source operand, which can be a register or memory location an immediate data with a destination operand that might be a register or a memory location. For the purpose of comparison, it subtracts the source operand from the destination operand but does not stock up the result anywhere. The flags are affected and depending on the result of the subtraction. If both of the operands are equal to zero flag is set. If the source operand is higher than the destination operand, carry flag is set or else is reset. The instance of this instruction are following:
Example :
1. CMP BX, 0100H Immediate
2. CMP 0100 Immediate [AX implicit]
3. CMP [5000H],OIOOH Direct
4. CMP BX, [SI] Register indirect
5. CMP BX, CX Register
General Bus Operation The 8086 has a joined data and address bus commonly referred to as a time multiplexed address and data bus. The major reason behind multiplexing address
SEG : Segment of a Label:- The SEG operator is which is used to decide the segment address of the, variable, label or procedure and substitutes the segment base address in plac
need help
code to add two matrices
Format of Control Register The format for the control register is given in Figure. Bit 0 of this register might be one before data may be output and bit two might be one
Write an Lc-3 assembly language program to read in a sequence of single-digit positive integers from the keyboard(one integer per line) until the sentinel value of 0 is reached and
init_lcd ;(this initialises a 2 row lcd) bcf TRISA,0 ;PORTA bit 0 as an output (lcd RS pin) bcf TRISA,1 ;PORTA bit 1
INC: Increment : - This instruction increments the contents of the particular memory or register location by the value 1. All the condition code flags are affected except the carry
Trying to convert small programs from C to 8086 assembly language using emu 8086 emulator. I converted to low level C, but struggling with converting to the Assembly language.
Program : A program to move a string of the data words from offset 2000H to offset 3000H the length of the string is OFH. Solution : For writing this program, we will use
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