Question

Asked Oct 17, 2019

NO 5

Step 1

The flow of the program is given as follows.

Mov ax,1750 ; 1750 is moved to ax.

Mov ds, ax ; Content in ax moves to ds that is 1750.

Mov si, 0100 ; location 0100 is moved to si.

Mov di, 200 ; location 0200 is moved to di.

Mov cx,0009 ; count 0009 is loaded in the register cx.

Inc cx ; value in the register cx is incremented by 1. now the value is 000a.

Inc si ; Memory location is incremented by 1. Now the location is 0101.

Mov al, [si+100H] ; location in SI is added with 100H that is 0200. To AL the content moved is '00' which is present in the location 0200. So now AX contains 1700.

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