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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
1. Write an assembly program that adds the elements in the odd indices of the following array. Use LOOP. What is the final value in the register? array1 DWORD 10, 20, 30, 40, 50, 6
ROR : Rotate Right without Carry: This instruction rotates the contents of destination operand to the bit-wise right either by one or by the count specified in register CL, exclud
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
Open notepad and enter the code for a program that calculates the following arithmetic expression: x = a + b + c - d - e + f The operands a, b, c, d, e, f, and x should be declared
can any one help me in my project by using assembly language
MLIL: Unsigned Multiplication Byte or Word: This instruction multiplies an unsigned byte or word by the contents of the AL. The unsigned byte or word can be in any one of the gene
what is the hex value in ax after executing the instructions ax= 1E8A bx=4080 add al,bl sub ah,bh
Assume that the registers are initialized to EAX=12345h,EBX =9528h ECX=1275h,EDX=3001h sub AH,AH sub DH,DH mov DL,AL mov CL,3 shl DX,CL shl AX,1 add DX,AX
A good starting point for your program is the toupper.asm program shown in class. It already queries the user for input and sets up a loop that looks at each character of the input
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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