Question

Translate HLL statement given below to Assembly

for (j = 0; j <= 100; j = j+1)

a = a + c

                              

Answer 1

Here eax register will store some value to which ecx will be added 100 times (same as doing a = a + c in HLL).

Firstly eax and ecx are initialized to some values (say 5 and 10 respectively) . Then inside the loop ecx will be added to eax 100 times using cmp.

mov eax,5

mov ecx,10

Loop:

add eax,ecx

cmp ecx,100

jle Loop

ret