3. Consider the following HCS12 Assembly Language code: CON EQU $10 DATA1 FCB$80

Question

3. Consider the following HCS12 Assembly Language code: CON EQU $10 DATA1 FCB$80 DATA2 FCB$50 ORG $0080 ; data declared here ORG $CO00 LDAADATA1 ; instructions start here SUBA DATA2 SUBA #CON a. Where are the bytes of data DATA1 and DATA2 located in memory? b What is the memory address of the first instruction? c. Does DATA1 have an initial value? If so, what is it? d. What is the significance of the # in front of CON in the third instruction? been executed? lines? e. What is in register A after the three instructions have f. What is the significance of the semicolon (G) in some

Explanation / Answer

a) DATA1 FCB $80

FCB means Fixed Constant Byte , so DATA1 is located at 80

and DATA2 located at 50

b) org $C000 i.e programming intialized from $C000

so the first instruction is started from $C000

c) in any programming DATA1 having some initialized value then only it will take memory from registers. here also we are having some value that is some GARBAGE value.

d) '#' is used for END OF TEXT.. i.e the program is end.

e) the value in DATA2 is filled in register A . i,e garbage value or any other value

f) ( ; ) is used for escape the line where it is mentioned.

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