《CS:APP》第二版第三章家庭作业部分答案

《CS:APP》第二版第三章家庭作业部分答案_第1张图片
<深入理解计算机系统>

以下答案均有个人完成转载注明出处

3.55

movl12(%ebp), %esiget x_l

movl20(%ebp), %eaxget y

movl%eax, %edx 

sarl$31, %edxget sign

movl%edx, %ecx 

imull%esi, %ecxcompute x_i_sign = sign * x_l

movl16(%ebp), %ebxget x_h

imull%eax, %ebxcompute dest_h = x_h * y

addl%ebx, %ecxcompute dest_h = x_h * y + x_l * sign

mull%esicompute 

  x_l * y

leal(%ecx, %edx), %edxcompute x_h * y + sign * x_l + sign

movl8(%ebp), %ecxget dest

movl%eax, (%ecx)dest_l = x_l * y

movl%edx, 4(%ecx)dest_h = ( x_h * y + sign * x_l ) + sign

大致算法

取底32位加后缀l, 取高32位加后缀h, sign为符号位扩展32位的y

dest_l =( x_l * y ) 取低位

dest_h = ( sign * x_l + x_h * y )_l + (x_l *y )_h


3.56

int loop ( int x, int n)

{

       int result = 1431655765;

       intmask;

       for(mask = 1 << 31 ; mask != 0 ; mask = (unsigned) mask >> n) {

              result ^= (x & mask) ;

       }

       returnresult;

}

A %esi x

 %edx mask

 %ebx n

 %edi %eax result

B result = 1431655765

 mask = -2147483648

C mask!=0

E 同上

F 填了


3.58

typedef enum { MODE_A, MODE_B, MODE_C,MODE_D, MODE_E } mode_t;

int switch3( int *p1, int *p2, mode_taction)

{

       intresult = 0;

switch ( action) {

case MODE_A; result = *p1; *p1 = *p2; break;

case MODE_B; result = *p2; *p2 = *p1; break;

       caseMODE_C; *p2 = 15; result = *p1; break;

       caseMODE_D; *p2 = *p1; result = 17; break;

       caseMODE_E; result = 17; break;

default: result= -1; break;

}

return result;

}


3.59

int switch_prob(int x, int n)

{

       intresult = x;

       switch(n){

              case40: case 42: result <<= 3; break;

              case43: result >>= 3; break;

              case44: result <<= 3; result -= x;

              case45; result *= result;

              default: result += 17;

       }

       returnresult;

}


3.60

int A [R] [S] [T]

A: 将等式从二维扩展到三维

A [ I ] [ j ] [ k ] = * ( A + 4 * ( i * S *T + j * T + k ) )

B:

R 11 S 7 T 9


3.62

A M: 76 / 4 = 19

B %edi i, %ecx j

C

void transpose ( int A [M][M] ) {

       int I, j;

       for( i = 0; I < M; i ++ )

              int*a = &A[0] [i];

              int*b = &A[i] [0];

              for( j = 0; j < I ; j++ )

                     int t = *a;

                     *a= *b;

                     *b= t;

                     b++;

                     a+= M;

              }

}



3.63

确定E1 和E2的定义

E1(n) = 3 * n

E2(n) = 2 * n – 1


3.64

A:

5行: result的address

6行: s1.v

7行: s1.p


B:

1: %ebp

2: s2.sum

3: s2.prod

4: s1.v

5: s1.p

6:返回地址

C向函数传递结构参数的通用策略: 将结构每一个元素当作本身的参数传入

D 从函数返回结构值的通用策略: 返回结构变量的初始地址作用, 因为没有寄存器可以存下一整个结构体, 且结构体的大小是可变的


3.66

A. 7

B typedef struct {

       intidx;

       intx [6];

} a_struct;

3.67

A

p: 0

x: 4

y: 0

next: 4

B: 这个结构一共需要8个字节

C:

void proc ( union ele *up )

{

       up -> e2.next -> e1.x = * ( up-> e2.next -> e1.p ) – up -> e2. y ;

}


3.69

A

long trace ( tree_ptr tp ) {

long ret = 0;

       while ( tp != NULL ) {

ret = tp.val;

       tp = tp.left;

       }

       returnret;

}

B 输出二叉树最左边的结点值


3.70

long traverse ( tree_ptr tp ) {

       longresult = 0;

       if( !tp ) return result;

       long VAL = tp->val;

       longleft = traverse ( tp->left );

       longresult = traverse ( tp->right );

       if( left < result )

              result= left;

       if( v < result )

              result= v;

       returnresult;

}

B 计算二叉树所有结点中的最小值

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