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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 ; 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;r4 <- r4+U3 L07: add r1, r1, 1 L08: jmp U4;goto U4 L09: end Which one of the following options is the CORRECT replacement for operands in the position (U1, U2, U3, U4) in the above assembly code?

1. (8, 4, 1, L02)
2. (3, 4, 4, L01)
3. (8, 1, 1, L02)
4. (3, 1, 1, L01)

Correct answer: (3, 4, 4, L01)

Solution

The correct option (3, 4, 4, L01) is right because the left shift operation in U1 must shift the value by 3 bits to multiply by 8 (2³), U2 should increment the address of array 'a' by 4 bytes (size of an int), and U3 should increment the address of array 'b' by 4 bytes as well. The jump back to L01 ensures the loop continues until the condition is met.

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