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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
8088 Timing System Diagram The 8088 address/data bus is divided in 3 parts (a) the lower 8 address/data bits, (b) the middle 8 address bits, and (c) the upper 4 status/
Write a program to evaluate the following expression. You are to evaluate the following equation: num1 - (input + num2) - (num3 + num4) Input will be a hex number input by
You are to create an assembly program for the MSP430 that correctly measures the wind direction, to a precision of 45° (N, NW, W, SW, S, SE, E, NE), using the MSP430's ADC. Your
Architecture Of 8088 The register set of 8088 is accurately the same as in to 8086. The architecture of 8088 is also same to 8086 except for 2 changes; a) 8088 has 4-byte instr
to separate positive and negative numbers
Mov ax, [1234h: 4336h + 100]
Intel's 8237 DMA controller : 1) The 8237 contain 4 independent I/O channels 2) It contains 27 registers, 7 of which are system-wide registers and 5 for each channel. 3)
DQ: Define Quad word:- This directive is taken in use to direct the assembler to reserve 4 words (8 bytes) of memory for the specified variable and can initialise it having
Modes of 8254 : Mode 0 (Interrupt on Terminal Count)-GATE which value is 1 enables counting and GATE which value is 0 disables counting, and GATE put not effect on
Program : Write a program to perform a one byte BCD addition. Solution : It is consider that the operands are in BCD form, but the CPU considers it as hexadecimal and acco
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