Question

Give the JAVA code for the following IJVM:

ILOAD j

ILOAD k

IADD

BIPUSH 3

I CMPEQ L1

ILOAD j

BIPUSH 1

ISUB

ISTORE j

GOTO L2

L1: BIPUSH 0

ISTORE k

L2:

Answer 1

• The given code simply storing two values in two variables
• Two variables are added and stored
• Then the sum is compared with the value 3, if the sum equals to 3 then the assigning k with 0, k=0
• If not subtracting 1 from j, j=j-1

Given code:

ILOAD j

ILOAD k

IADD

BIPUSH 3

I CMPEQ L1

ILOAD j

BIPUSH 1

ISUB

ISTORE j

GOTO L2

L1: BIPUSH 0

ISTORE k

L2:

Note: The given code has some corrections to make it work properly, so here is the updated code (Result is same but verify the code once for any correction. It is for student best understanding)

Updated Code:

ILOAD j //push j onto the stack

ILOAD k //push k onto the stack

IADD //pop j and k then add both

ISTORE i //i=j+k, store in i

ILOAD i //push i onto the stack

BIPUSH 3 //push 3 onto the stack

I CMPEQ L1 //compare 3 with i, if equal goto L1

ILOAD j //if i not equal with 3, push j onto the stack

BIPUSH 1 //push 1 onto the stack

ISUB //subtract 1 from j

ISTORE j //store j=j-1 in j, then push j onto the stack

GOTO L2 //go to L2

L1: BIPUSH 0 //if -==3, then push 0 onto the stack

ISTORE k //store 0 in k, then push onto the stack

L2: //end of the else loop

==================

Java code:

i = j + k;

if( i == 3)

k = 0;

else

j = j-1;

==================

Code cheats:

ILOAD variable: push local variable onto the stack

IADD: pop top two elements from stack then add, push the sum onto the stack

ISTORE variable: pop top element from stack and store it in variable

BIPUSH num: push a byte number onto the stack

I CMPEQ: compares top two elements from the stack

I CMPEQ L1: compares top two elements from the stack, if equal goto L1 else proceed with next step

ISUB: pop top two elements from stack then subtract them, push the subtract onto the stack

GOTO L2: tells to go to particular step, here it is L2

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