卷卷天下卷卷天下

1、Introduction to Computer Systems Fall 2007Midterm ExaminationNameStudent No.ScoreProblem 1: (22 points)Assume we are running code on a 6-bit machine using twos complement arithmetic for signed integers. A “short” integer is encoded using 3 bits.Fill in the empty boxes in the table below. The followi

2、ng definitions areshort sy = -2;unsigned short usy = sy int y = sy;int x = -31; unsigned ux = x;used in the table: (1*20 points)Bina Represe230110681012141618202221-TMin19TMax17UMax15x&!y13x+y11x19usy7y5ux4-9-1zeroDecimal RepresentationExpressionNote: You need not fill in entries marked with “-”.卷卷天

3、下卷卷天下Problem 2: (20 points)Fill the following functions. (1*20 points)/* isLess - if x ythen return 1, else return 0* Example: isLess(4,5) = 1.* Legal ops: ! & | + * Max ops: 24* Rating: 3*/int isLess(int x, int y)int temp1 = _1_; int temp2 = temp1 + 1; int temp3 = x + temp2;int expression1 = x_2_ t

4、emp1;int expression2 = temp3 _3_ ( ( x y );int ret = ( ( expression1 _4_ expression2 ) _5_ ) & _6_ ; return ret;Part1 (1*6 points)/* Returns 1 if x is a power of 2 and x 0, otherwise return 0* Examples: isPower2(5) = 0, isPower2(8) = 1, isPower2(0) = 0* Note that no negative number is a power of 2.*

5、 Legal ops: ! & | + * Max ops: 60* Rating: 4*/int isPower2(int x)int temp = x + _1_;int temp1 = ( x _2_ temp );int ret= ( _3_x ) & ( ! ( _4_ + temp1 ) ) & ( ( x 31 ) & 1 ); return ret;Part2 (1*4 points)/* Returns a count of the number of 1s in the argument.* Examples: bitCount(5) = 2, bitCount(7) =

6、3* Legal ops: ! & | + * Max ops: 40* Rating: 4*/int bitCount(int x)int common1 = ( _1_ 8) + _2_ ;int common = ( common1 _4_ ) & common ) +( x _5_ ) & common) +( x _6_ ) & common ); int temp1 = temp + ( temp _7_);intresult = ( temp1 & 0xf ) + ( temp1 _8_ ) & 0xf ) +( temp1 _9_ ) & 0xf ) + ( temp1 _10

7、_ ) & 0xf ); return result;Part3 (1*10 points)nc_switch:ushl %ebpovl %esp, %ebp bl $8, %espovw $4, -2(%ebp)notvbfu$n1c5_,s-w3(i%tcheb(ipn)t a, int b) ovshl o8r(t%se=bp), %1eax ; bcl h$a3r3c, =%eax2;ovl %eax, -8(%ebp)mpslw$it7c,h-(8(%eb3p) .cLa1s9e4:Problem 3: (30 points)Fills the blanks in source co

8、de and x86 assembly code for func_switch.Source code: (115 points)x86 assembly code: (35 points)ovl -s8(=%beb-p5);, %edxovl .L27(,5%edx,;4), %eaxp c*a%seeax6:Reecctihoenck.r:odatalign 4ign 47;8;7:case9: ong .bL2=0 0x7F;ong .aL2=2 0;ong .L2310;ongca.Lse1911:ong .bL1=9 a * 8;ongca.Lse1912:ong .bL2=4 a

9、 0xF; ongde.Lfa2u5lt:ext13; 0:ovl 121(%4ebp),; %eax return_1 15;_2 p .L192:ovsbl -3(%ebp),%eax ovl %eax, 12(%ebp) p .L193:ovl $127, 12(%ebp)_ 3 p .L22 4:ovl 8(%ebp), %eax4 ovl %eax, 12(%ebp) 5:ovl 8(%ebp), %eax5 ovl %eax, 12(%ebp) 9:ovw -2(%ebp), %dxProblem 4: (28 points)Consider the following assem

10、bly representation of a function testtest:pushl %ebpmovl %esp, %ebp pushl %esipushl %ebx subl $16, %espmovl $0, -16(%ebp) movl $0, -20(%ebp).L6:movl -20(%ebp), %eax cmpl 12(%ebp), %eax jl.L9jmp.L7.L9:movl -20(%ebp), %eax leal0(,%eax,4), %edx movl 20(%ebp), %eax movl $0, (%eax,%edx) movl $0, -24(%ebp

11、).L10:movl -24(%ebp), %eax cmpl 16(%ebp), %eax jl.L13jmp.L8.L13:movl -20(%ebp), %eax imull 16(%ebp), %eax addl -24(%ebp), %eax leal0(,%eax,4), %edx movl 8(%ebp), %eax movl (%eax,%edx), %eax movl %eax, -12(%ebp) movl -20(%ebp), %eax leal0(,%eax,4), %ebx movl 20(%ebp), %esi movl -20(%ebp), %eax leal0(

12、,%eax,4), %ecx movl 20(%ebp), %edx movl -12(%ebp), %eax addl (%edx,%ecx), %eaxmovl %eax, (%esi,%ebx) movl -12(%ebp), %eax cmpl -16(%ebp), %eax jle.L12movl -12(%ebp), %eax movl %eax, -16(%ebp).L12:leal-24(%ebp), %eax incl(%eax)jmp.L10.L8:leal-20(%ebp), %eax incl(%eax)jmp.L6.L7:movl -16(%ebp), %eax ad

13、dl $16, %esppopl %ebx popl %esi leaveret(1)Assume right after going into the function test (i.e., before the instruction “pushl %ebp”), the value of %ebp is 0xbffff9A4 and the value of %esp is 0xbffff62C. Fill the following table with the values stored at the indicated memory addresses and registers

14、 right after the last instruction of this function. If you do not know the value of some address, just writeAddressValue0xbffff61010xbffff61420xbffff61830xbffff61C40xbffff62050xbffff62460xbffff62870xbffff62C0x804592c0xbffff6300x804A92c0xbffff6340x40xbffff6380x30xbffff63C0x804A9640xbffff6400x804A9300

15、x 804A97080x 804A96C90x 804A968100x 804A964110x 804A960120x 804A95C0x80x 804A9580x250x 804A9540x1D0x 804A9500x470x 804A94C0x350x 804A9480x2E0x 804A9440x200x 804A9400x160x 804A93C0xb0x 804A9380x80x 804A9340x230x 804A9300x180x 804A92c0x10down “*” and more reason is preferred (1*12 points)LabelThe nute

16、st.L6.L9.L10.L13.L12.L8.L7(2)How many times does this function read or write the value of certain memory address when executing with the given values? (Assume executing incl (%eax) once, referencing memory twice). Note: Explain the procedure asdetailed as possible, for example, how many times does e

17、ach instruction execute and you can mark it on the right ofthe instruction and show why, and furthermore, you can list the numbers of these 8 label, such as, .L6 * and so on. (28 points)Solution:DecimalRepresentationBinaryProblem 1: (22points 1*22)Problem 2: (20points)1.(1*6points)(1)(2)(3)(4)(5)(6)2. (1*4 points)(1)(2)(3)(4)3