The following statements will show a few possible ways of using Label, Instruction and Comments. Assembly Language Directives are not the instructions to the Microcontroller Assembler even though they are written in the Mnemonic field of the program.
Assembly Language Directives are actually instructions to the Assembler and directs the Assembler Program what to do during the process of Assembling. We will now see about few of the important and frequently used Assembly Language Directives. This is also the address from which the Microcontroller will start executing the code.
Jump to MAIN. MAIN starts here. The DB Directive is used to define a Byte type variable. Only lower byte is. The END Directive is used to stop the assembling process. This should be the last statement in the program. In this tutorial, we have seen about the basics of Microcontroller Assembly Language Programming, the Structure of Assembly Language for , Assembly Language Directives and few examples.
The CPU can access the data in different ways by using addressing modes. The microcontroller consists of five addressing modes such as:. It is used for immediately storing the value in the memory registers. In this addressing mode, the source and destination must be a register, but not general purpose registers. So the data is not moved within the general purpose bank registers. In this addressing mode, the source or destination or both source and destination must be an address, but not value.
In this addressing mode, the source or destination or destination or source must be a indirect address, but not a value. This addressing mode supports the pointer concept. The pointer is a variable that is used to store the address of the other variable. This pointer concept is only used for R0 and R1 registers. This addressing mode is used to read the data from the external memory or ROM memory. All addressing modes cannot read the data from the code memory.
The code must read through the DPTR register. The DPTR is used to point the data in the code or external memory. The instruction set is the structure of the controller or processor that provides commands to the controller to guide the controller for processing data.
The instruction set consists of instructions, native data types, addressing modes, interrupt registers, exceptional handling and memory architecture. In case of the programming different types of CISC instructions include:.
The CPU executes the instructions based on the condition by checking the single bit status or byte status. With these tools, you can generate embedded applications for the multitude of derivatives. Keil provides following tools for development.
C51 Optimizing C Cross Compiler,. A51 Macro Assembler,. Example generate a square wave of 10 Hz at pin P1. Example write a program to make 8 to 1 multiplexer with enable signal. Example take parallel input from port P1 convert it into serial and send it via P0.
Subb a, r6 ; subtract it with other. Mov 20h, a ; store the result. Mov a, r5 ; get the first higher byte. Subb a, r7 ; subtract from other. Mov 21h, a ; store the higher byte. Statement 4: — divide the content of r0 by r1. Store the result in r2 answer and r3 reminder. Then restore the original content of r0. Solution:- after getting answer to restore original content we have to multiply answer with divider and then add reminder in that. Mov a, r0 ; get the content of r0 and r1. Mov b, r1 ; in register A and B.
Div ab ; divide A by B. Mov r2, a ; store result in r2. Mov r3, b ; and reminder in r3. Mov b, r1 ; again get content of r1 in B. Mul ab ; multiply it by answer. Add a, r3 ; add reminder in new answer. Mov r0, a ; finally restore the content of r0. Statement 5: — transfer the block of data from 20h to 30h to external location h to h. Solution: — here we have to transfer 10 data bytes from internal to external RAM.
So first, we need one counter. Then we need two pointers one for source second for destination. Mov r7, 0Ah ; initialize counter by 10d. Mov r0, 20h ; get initial source location.
Mov dptr, h ; get initial destination location. Nxt: Mov a, r0 ; get first content in acc. Movx dptr, a ; move it to external location. Inc r0 ; increment source location. Inc dptr ; increase destination location. Djnz r7, nxt ; decrease r7.
Statement 6: — find out how many equal bytes between two memory blocks 10h to 20h and 20h to 30h. Solution: — here we shall compare each byte one by one from both blocks. Increase the count every time when equal bytes are found. Mov r0, 10h ; get initial location of block1.
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