Csapp (deep understanding of computer Systems) Second Edition homework answer-chapter III __ Computer

The answer is by oneself completes, and experiment or debugging, welcome the reference.

3.54 int decode2 (int x,int y,int z) {int t=z-y;
	int t2=t<<15;
	t2=t2>>15;
Return (x^t) *t2; 3.55 MOVL (%EBP),%esi; The low movl of Get X (%EBP),%eax; Get y movl%eax,%edx sarl $31,%edx; the sign bit movl%edx,%ecx the imull;%esi,%ecx (0 or-1)%ecx=x (*y), MOVL; High Imull%eax,%ebx of Get x%ebx=x high *y addl%ebx,%ecx high%ecx=x low *y+x sign bit (0 or-1) *y Mull,%esi (x low%edx=) high *y x= (x's low *y) low Leal (%ecx,%edx),%edx MOVL 8 (%EBP),%ecx movl%eax, (%ECX) movl%edx,4 (%ECX) (x's low *y) high +x's low *y+x sign bit ( 0 or-1 low position stored at high (x low *y) in Low (Xl*y) _h+xh*y+xl*-1 (x1*y) _l xl=1000 xh=0000 y=1001 (x*yi=x*) yu-16 equivalent high = (=x*yu-16x) _ H-x, low = (X*yu) _l because this is mull unsigned multiplication 3.56 a. X:%esi,n:%ebx,result:%edi,mask:%edx B. result=0x55555555;
0101 0101 0101 0101 0101 0101 0101 0101;
	C. mask!=0 D. Mask Each loop to the right N-bit E. Result Each loop and x&mask the value of a different or F. int loop (int x,int n) {int result=0x55555555;
	int mask; for (mask=0x80000000;mask!=0 (unsigned) mask) >>n)//assembly mask is the logical right SHIFT, and mask is thereThe symbolic integral type should be converted to unsigned integral type to ensure that the right shift operator of the C language is compiled into the logic of the assembly to move right {result^=x&mask;
return result;
	3.57 int Cread_alt (int *xp) {int t=0;
	int *p=xp?xp:&t;
return *p;
	3.58 int Switch3 (int *p1,int *p2,mode_t action) {int result=0;
			Switch (action) {case MODE_A:RESULT=*P1;
			*P1=*P2;
		Break
			Case MODE_B:RESULT=*P1+*P2;
			*p2=result;
		Break
			Case Mode_c: *p2=15;
			RESULT=*P1;
		Break
		Case Mode_d: *P2=*P1;
			Case mode_e:result=17;
		Break
			Default:result=-1;
	Break
return result;
	3.59 int switch_prob (int x,int n) {int result=x;
			Switch (n) {case 40:case 42:result<<=3;
		Break
			Case 43:result>>=3;
		Break
			Case 44:result<<=3;
		Result-=x;
		Case 45:result*=result;
			Case 41://Can not default:result+=11;
	Break
return result;
} 3.60 A. A[i][j][k]=a+ ((2^6-1) *i+9j+k) *4=a+ (63i+9j+k) *4 B. t=9 s*t=63 r*s*t=2772/4=693 r=11 s=7 t=9 3.61 int var_prod_ele (int n, I NT A[n][n], int b[n][n], INT I, int k) {int result = 0;
    int *a = &A[i][0];
    int *b = &B[0][k];
    int *e = &A[i][n];
    for (; a!=e;)
        {result = *a * *B;
        B+=n;
    a++;
return result;
The following is the looping part of its assembly code: EDI is 4*N,EBX and edx respectively is B and A,esi is E,eax is result.
ECX is a register that is used to store multiplication. L4:MOVL (%EBX),%ecx imull (%edx),%ecx addl%ecx,%eax addl%edi,%ebx addl $,%edx cmpl%edx,%esi jne L4, Result,arow, Bcol,4*n these four values of memory required to convert the value of N to E to consume a register to compare the value of N and a, save the variable J, use a register less 3.62 a. m=76/4=19 B. i:%edi,j:ecx C. void Transpose (int A M
		[M]) {for (int i=0;i<m;i++) {int *arr=&a[i][0];
		int *brr=&a[0][i];
			for (int j=0;j<i;j++) {int t=*arr;
			*ARR=*BRR;
			*brr=t;
			Arr+=1;
		Brr+=m; }} 3.63 #define E1 (n) 3*n #define E2 (n) 2*n-1 3.64 A. 8 (%EBP) &s2 (%EBP) s1.p (%EBP) s1.v B.------------%eb P s2.sum s2.prod s1.v s1.p &s2------------%esp C. Passes the values of each member of the struct to method D. The caller passes the address of the return value variable to the method 3.65 a=3 b=7 3.66 the A.CNT=196/28
	=7 B. typedef struct {int idx; int x[6];
}a_struct; ap->idx=* (bp+1c*i+4) *4 ap->x[ap->idx]= (bp+4+28*i+4) +* (bp+0x1c*i+4) *4=bp+8+28i+* (bp+28i+4) *4=bp+8+4 (7i+ * (bp+28i+4)) 3.67 A. e1.p:0 e1.x:4 e2.y:0 e2.next:4 b.8 C. VOID proc (Union ele *up) {up->e2.next->e1.x=* (up->e
2.NEXT-&GT;E1.P)-up->e2.y;
    } 3.68 void Good_echo () {char C;
    int x = 0;
    while (X=getchar (), x!= ' \ n ' && x!=eof) {putchar (x);
	} 3.69 A. Long Trace (tree_ptr tp) {long result=0;
		while (TP) {result=tp->val;
	tp=tp->left;
return result; B. Output the value of the leftmost node of the binary tree 3.70 A. Long traverse (tree_ptr tp) {if (!TP) return 9223372036854775807;//long.max long V=tp->val
	;
	Long Left=traverse (tp->left);
	Long Result=traverse (tp->right);
	if (Left<result) result=left;
	if (result>v) result=v;
return result; B. Find the smallest value in a binary tree node

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