Discussion
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\(\lt\)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 \(\lt\)- Memory[r4+0]
L03: shl r5, r5, U1 ;r5 \(\lt\)- r5 \(\lt\lt\) U1
L04: sw r5, 0(r3) ;Memory[r3+0] \(\lt\)- r5
L05: add r3, r3, U2 ;r3 \(\lt\)- 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?
Answer: B
The operands are (3, 4, 4, L01).
- U1: multiplying by 8 is a left shift by 3 bits, since \(8 = 2^{3}\). A shift of 8 would multiply by 256.
- U2 and U3: an int is 32 bits, 4 bytes, and memory is byte-addressable, so the pointers r3 (into a) and r4 (into b) must each advance by 4 to reach the next element.
- U4: after r1 is incremented, the loop must test r1 == r2 again before the next iteration, so it jumps back to L01. Jumping to L02 skips the test and never leaves the loop.
(8, 4, 1, L02) gets the shift, the b pointer and the jump wrong; (8, 1, 1, L02) also steps the pointers by a single byte; and (3, 1, 1, L01) has the right shift and jump but moves only one byte per element, reading and writing overlapping parts of the integers.
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