In 8085 Instruction set, this instruction MOV M, r will copy 8-bit value from the register r to the memory location as pointed by HL register pair.This instruction uses register addressing for specifying the data.
As “r” can have any one of the seven values −
r = A, B, C, D, E, H, or L
Thus there are seven opcodes for this type of instruction. It occupies only 1-Byte in memory.
|Mnemonics, Operand||Opcode(in HEX)||Bytes|
|MOV M, A||77||1|
|MOV M, B||70||1|
|MOV M, C||71||1|
|MOV M, D||72||1|
|MOV M, E||73||1|
|MOV M, H||74||1|
|MOV M, L||75||1|
MOV M, E is an example instruction of this type. It is a 1-Byte instruction. Let us suppose, E is having the initial value ABH, HL register pair is pointing to the memory location 4050H, 4050H memory location’s content is CDH. Then after execution of the instruction MOV M, E, E register’s content will be CDH. The result of execution of this instruction is shown with this example is shown below −
|2005||273||MOV M, E||Memory location pointer by HL register pair <- E|
Here is the timing diagram of the instruction MOV M, E as below.
Summary − So this instruction MOV M, E requires 1-Byte, 2-Machine Cycles (Opcode Fetch, Memory Read) and 7 T-States for execution as shown in the timing diagram.