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GATE CSE 2023 Q41: Fill operands U1–U4 in assembly for a[i] = b[i] * 8

Question

Consider the given C-code and its corresponding assembly code, with a few operands U1–U4 being unknown. Some useful information as well as the semantics of each unique assembly instruction is annotated as inline comments in the code. The memory is byte-addressable.

//C-code
int a[10], b[10], i;
// int is 32-bit
for (i=0; i<10;i++)
    a[i] = b[i] * 8;

;assembly-code (; indicates comments)
;r1-r5 are 32-bit integer registers
;initialize r1=0, r2=10
;initialize r3, r4 with base address of a, b

L01: jeq r1, r2, end  ;if(r1==r2) goto end
L02: lw  r5, 0(r4)    ;r5 <- Memory[r4+0]
L03: shl r5, r5, U1   ;r5 <- r5 << U1
L04: sw  r5, 0(r3)    ;Memory[r3+0] <- r5
L05: add r3, r3, U2   ;r3 <- r3+U2
L06: add r4, r4, U3
L07: add r1, r1, 1
L08: jmp U4           ;goto U4
L09: end

Which one of the following options is a CORRECT replacement for operands in the position (U1, U2, U3, U4) in the above assembly code?

Try it: multiple choice: exactly one correct option

Select one option
Wrong answer costs 2/3 mark in GATE.
Show answer and explanation

Answer: B (official GATE 2023 final answer key)

Concept used: Multiplying by 2k2^k is a left shift by kk; array pointers advance by the element size in bytes.

First step: Write 8 as a power of 2 to get U1, and use the 4-byte int size for the pointer increments.

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Source: official GATE 2023 CS question paper and final answer key, organised by IIT Kanpur. Question paper · Answer key

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