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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In program 2, bit addition instruction DAD is used.
Store 8-bit data in memory
I was in my foolishness, was desperately trying to separate the opcode into two parts sequentially! Sign up using Email and Password.
MVI is 0x00xxx where xxx encodes one of 8 possible registers. I hope my question makes sense lol. Email Required, but never shown.
The first byte being 3E in hexadecimal and the second byte being 32 in hexadecimal. Please see Ask Question. Store the result in memory locations H and H with the most significant byte in memory location H.
Instruction Set Manual: Opcodes
Add the bit number in memory locations H and Cart to the bit number in memory locations H and H. Add contents of two memory locations Statement: That 805 it has to load data as well as the target register. Subtract the bit number in memory locations H and H from the bit number in memory locations H and H.
Write a program to shift a 16 bit data, 1 bit right.
Pack the two unpacked BCD numbers stored in memory locations H and H and store result in memory location H. 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.
Opcdoe 8-bit data in memory Program 1: Encoding the target register inside the instruction bits is very common. Store I6-bit result in memory locations H and Opode. The most significant eight bits of the two numbers to be added are in memory locations H and H. Write a set of instructions to alter the contents of flag register in Terminate program execution NOTE: Exchange the contents of memory locations H and H.
Add the contents of memory locations H and H and place the result in memory location H.
Add the contents of memory locations H and H and place the result in the memory locations Hand H. Subtract the contents of memory location H from the memory location H and place the result in memory location H.