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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
Machine Level Programs In this section, a few machine levels programming instance, rather then, instruction sequences are presented for comparing the 8086 programming with that
can u please give me ideas on Assembly Language Projects using Nasm
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Memory Address Decoding Binary Decoders - Decoders have 2n-inputs and n outputs, each input combination results in a single output line contain a 1, and all other lines contain
Register Organization of 8086 8086 has a great set of registers containing special purpose and general purpose registers. All the 8086 resisters are 16-bit registers.
How to write an assembly program The initial step in writing an assembly language program is to identify and study the problem. After studying the problem, choose the logical m
Assembly Code for Reading Flow & Generating Serial Output The timer is timer 1 is set for the baud rate 9600, as the crystal used is of 11.0592 Hz. Then the timer 1 is starte
Signal descriptions of 8086 : described below are common for the maximum andminimum mode bothdata lines AD15 -AD0: These are the time multiplexed andmemory I/O address. Addre
Compute the Fibonacci sequence - assembly program: Problem: Fibonacci In this problem you will write a program that will compute the first 20 numbers in the Fibonacci sequ
Memory Segmentation : The memory in an 8086/8088 based system is organized as segmented memory. In this scheme, the whole physically available memory can be divided into a n
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