a. f = f + g + h + i + j + 2; b. f = g - (f + 5); a) For the C statements above,
ID: 3616681 • Letter: A
Question
a. f = f + g + h + i + j + 2; b. f = g - (f + 5); a) For the C statements above, what is the corresponding MIPSassembly code? Use a minimal number of MIPS assemblyinstructions. b) For the C statements above, how many MIPS assemblyinstructions are needed to perform the C statement? c) If the variables f, g, h, i, and j have values 1, 2, 3, 4,and 5, respectively, what is the end value of f? a. f = f + g + h + i + j + 2; b. f = g - (f + 5); a) For the C statements above, what is the corresponding MIPSassembly code? Use a minimal number of MIPS assemblyinstructions. b) For the C statements above, how many MIPS assemblyinstructions are needed to perform the C statement? c) If the variables f, g, h, i, and j have values 1, 2, 3, 4,and 5, respectively, what is the end value of f?Explanation / Answer
a) add t0, f, g #temporary variable t0contains f+g
add t1, h, i #temporary variablet1 contains h+i
addi t2, j, 2 #temporary variable t2 containsj+2
add t1, t1, t2 #t1 contains t1+t2 which is (h + i)+(j + 2)
addi t2, j, 2
add t1, t1, t2
add f, t0, t1
f = f + g + h + i + j +2
t0 = f + g = 1 + 2 = 3
t1 = h + i = 3 + 4 = 7
t2 = j + 2 = 5 + 2 = 7
t1 = t1+ t2 = 7 + 7 =14
f =t0 + t1 = 3 + 14 = 17
sub f, g, t0
f = g - (f + 5)
t0 = f + 5 = 1 + 5 = 6
f = g- (f+5) = 2 – (1+ 5)
f = -4
Therefore the final value of f is -4
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.