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D) There are three values possible after the execution of both
processes. These are 14,15, 16.
Explanation:
![Case 1:- (heks consider initial value of x 15 ) #idding 1 LDAP.1/ ADD PS, 1. STA RIX LDARAH EUR 21, 2 & seebtracting 1 STA B1](//img.homeworklib.com/questions/b42bfc70-9956-11eb-9879-19a0ef4bc10f.png?x-oss-process=image/resize,w_560)
Case 1:- (heks consider initial value of x 15 ) #idding 1 LDAP.1/ ADD PS, 1. STA RIX LDARAH EUR 21, 2 & seebtracting 1 STA B1, In this case the value of x at the end would be 15. First Pg executes all the statements, x becomes 16. Then Pexecutes all the statements and X becomes 15. Case 2: PA P2 LDA R2, X I LDA R1, X ADD P1, 1 SUB R2,1 STA 1,5 STA R2, X In this case the value of x at the end rwould be 14. Here when Pa loods x into R1, value of x is 15. And at the end P2 stores from R 1 to x. Hence value of x is 14. Case 3:0 - PA LDA R2, X LDA P. 2,4 SUB R.1, 1 ADD R1, 1 STARIX STA R2, X P2 - In this case value of x at the end would be 16. When l access the value of x its value is 15. Pg stores the at the end so, its incremented value would be stored