Q.Sample of Assembly Language Program: explain the following 8051 assembly language program, which loads two numbers and accumulates them (with 10H) in the accumulator.

ORG OH          ; start (origin) at location 0  
MOV R2, #15H    ; load 15H into R2  
MOV R6, #23H    ; load 23H into R6  
MOV A, #0       ; load 0 into A  
ADD A, R2       ; add contents of R2 to A ; now A = A + R2  
ADD A, R6       ; add contents of R6 to A ; now A = A + R6  
ADD A, #10H     ; add to A value 10H ; now A = A + 10H  
HERE: SJMP HERE ; stay in this loop  
END             ; end of ASM source file  

The program clears A and then adds the contents of R2 (15H) and R6 (23H) and the immediate value 10H into it, so A holds the running sum; ORG 0H places the code at address 0 and END marks the end of the source.

[!ANSWER]

A = 15H + 23H + 10H = 48H, after which the program parks in the infinite loop HERE: SJMP HERE.

[!TLDR]

The program initialises A to 0 and adds 15H, 23H and 10H to it, leaving A = 48H, then loops in place.

Concept - directives vs instructions. An assembly program mixes instructions (MOV, ADD, SJMP - commands executed by the CPU) with directives (ORG, END - directions to the assembler, not the CPU). ORG 0H tells the assembler to place the following code starting at program-memory address 0; END marks the end of the source file.

Line-by-line:

• MOV R2, #15H - load 15H into R2.
• MOV R6, #23H - load 23H into R6.
• MOV A, #0 - clear the accumulator, A = 0.
• ADD A, R2 - A = 0 + 15H = 15H.
• ADD A, R6 - A = 15H + 23H = 38H.
• ADD A, #10H - A = 38H + 10H = 48H.
• HERE: SJMP HERE - an unconditional short jump to its own label, holding the processor in an infinite loop so it does not run past the program.

Result: 15H + 23H + 10H = 21 + 35 + 16 = 72 decimal = 48H.

[!ANSWER]

A = 48H (72 decimal); the program then stays in the loop HERE: SJMP HERE.

[!NOTE]

The textbook typesets the hexadecimal digit 0 (zero) as the capital letter "O" - the first line prints as "ORG OH", but the correct 8051 directive is ORG 0H.