Each instruction has a one-byte (8-bit) operation codes or opcode. With 8- bit binary opcode, a total of different operation codes can. Intel instruction set. x0, x1, x2, x3, x4, x5, x6, x7, x8, x9, xA, xB, xC, xD, xE, xF. 0x, NOP 1 4 , LXI B,d16 3 10 , STAX B 1 7 , INX B 1 6 –K 1 1 ADDRESSING MODES OF Shown in the following are the sizes of a 5CH This can be verified from the opcode chart given in the previous chapter.
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So the opcode if it can be broken down sometimes they dont break down into smaller parts, depends on the instruction set would have to be in that first byte. Write a program to shift a 16 bit data, 1 bit right. Add the bit number in memory locations H and H to the bit number in memory locations H and H.
Subtract the contents of memory location H from the memory location H and place the result in memory location H. In program 2, bit addition instruction DAD is used. I hope my question makes sense lol.
Write a set of instructions to alter the contents of flag register in The first byte being 3E in hexadecimal and the second byte being 32 in hexadecimal. Exchange the contents of memory locations Statement: The color coding on that chart gives a strong indication of the opcode decoder if the 2 msbits are 00 then if the lower 2 bits are 10 then if bit 2 is a 1 then it is an MVI and bits determine which register. How does “3E” tell the microprocessor both the information?
Add contents of two memory locations Statement: Pack the two unpacked BCD numbers stored in memory locations H and H and store result in memory location H. Assume the least significant digit is stored at H.
Store 8-bit data in memory
Add the contents of memory locations H and H and place the result in the memory locations Hand H. I was in my foolishness, was desperately trying to separate the opcode into two parts sequentially! Assume that data is in BC register pair. In program 1 direct addressing instruction is used, whereas in program 2 indirect addressing instruction is used. Read the program given below and state the contents of all registers after the execution of each instruction in sequence.
Lpcode the contents of memory locations H and H. MVI is 0x00xxx where xxx encodes one of 8 possible registers. My book says that it is a two byte instruction where the first byte is the opcode and the second is the operand.
Two digit BCD number is stored in memory location H.
Instruction Set Manual: Opcodes
Add two 8-bit numbers Statement: Sign up using Email and Password. Sign up or log in Sign up using Google.
Add the contents of memory locations H and H and place the result in memory location H. Post as a guest Name.