CS 33 Mid Term ExamUID : Name: Fill in your answers below the questions. If you need more space, you may usethe back of the page but be sure to indicate on the front that you are doingso. Be sure to label multi-part answers. All answers are worth 10 points each.1. Problem # 2.71, CSPP page 124. Topic: type conversion A. The return expression yields an unsigned result. The specificationsays that each packed byte is intended to be signed. B. Either a cast is required or setting the expression into a char and returning that instead would work. char a = ( word >> ( bytenum << 3 ) ) & 0xff ; return (int) a ; or int left = word << ((3-bytenum) << 3); return left >> 24; 2. Problem # 2.76, CSPP page 126. Topic: multiply by a constant using shiftand add A. k = 17 ( x << 4 ) + x 16*x+x B. k = -7 x -( x << 3 ) x-8*x C. k = 60 ( x << 6 ) - ( x << 2 ) 64*x-4*x D. k = -112 ( x << 4 ) - ( x << 7 ) 16*x-128*x CS 33 Mid Term Exam - 1 -3. Problem # 2.81, CSPP page 126-127. Topic: Boolean operations Part A 32 A. (x<y) == (-x>-y). No, Let x = TMin , y = 0. B. ((x+y)<<4) + y-x == 17*y+15*x. Yes, from the ring properties of two’s complement arithmetic. C. ˜x+˜y+1 == ˜(x+y). Yes, ˜x+˜y+1 = (!x!1)+(!y!1)+1 = !(x+y)!1 = ˜(x+y). D. (ux-uy) == -(unsigned)(y-x). Yes. Due to the isomorphism between two’s complement and unsigned arithmetic. E. ((x >> 2) << 2) <= x. Yes. Right shift rounds toward minus infinity. 4. Problem # 2.87, CSPP page 129. Topic: Floating point decoding Format A Format BBits Value Bits Value1 01111 001 -9/8 1 0111 0010 -9/80 10110 011 176 0 1110 0110 1761 00111 010 -5/2 1 0000 0100 -5/210 100 00000 111 7/2 0 0000 1110 7/217 101 11100 000 -2 1 1110 1111 -248130 10111 100 384 0 1110 1111 infinity due roundoffCS 33 Mid Term Exam - 2 -5. Problem #s 2.92-2.95, CSPP page 132-133. Topic: Bit level float pointFor these 5 mini problems,simply write the missing line in the function belowas your answer. You do not need to worry about the last paragraph in theproblem: what to do in abnormal cases and test all possible values.The function is:unsigned do_the_problem(unsigned f) { /* Decompose bit representation into parts */ unsigned sign = f>>31; unsigned exp = f>>23 & 0xFF; unsigned frac = f & 0x7FFFFF; *** your one line statement would go here, just put it in the answer line *** /* Reassemble bits */ return (sign << 31) | (exp << 23) | frac; }Prob Answer2.92 sign = 0 ;2.93 exp = exp+1 ;2.94 exp = exp-1 ;2.95 **problem misstated**6. Problem # 3.58, CSPP page 297-298. Topic: Interpreting assembly codeMODE_A result = *p1 ; *p1 = *p2 ; break ;MODE_B result = *p2 ; *p2 = *p1+*p2 ; break ;MODE_C *p2 = 15; result = *p1 ; break ;MODE_D *p2 = *p1 ; MODE_E result = 17 ; break ;default result = -1 ;CS 33 Mid Term Exam - 3 -7. Problem # 3.60, CSPP page 300. Topic: How arrays are laid out in memory A. &A[0][0][0]+4*(i*S*T+j*T+k) B. A[5][9][11] 8. Problem # 3.62, CSPP page 300-301. Topic: How arrays are laid out inmemory. *** Do not do part C. *** A. M = 13 B. I is %edi, j is %ecx CS 33 Mid Term Exam - 4 -9. Problem # 3.65, CSPP page 304. Topic: How structures containing arrays arelaid out in memory the offset to str1.y is 92 but there could be padding. with padding A*B*2 could be 90 which would require padding the offset to str2.u minus the offset to str2.t minus 4 would be the sizeof s.but there could be padding so B is either 9 or 10. if it is 9 there would be padding in both and with A=5 and B=9, A*B*2 = 90 andthere would be padding in str1. so A=5, B=9. 10. Problem # 3.67, CSPP page 305. Topic: Unions and structures. Use IA32pointer sizes. A. e1.p offset 0; e1.y offset 4; e2.x offset 0; e2.next offset4 B. 8 bytes C. up -> e2.next -> e1.y = *( up -> e2.next -> e1.p ) - up -> e2.x CS 33 Mid Term Exam - 5