Homework 5 & 6

1100012749

---

---

3.61

int var_prod_ele(int n, int A[n][n], int B[n][n], int i, int k) {
   void *Arow = (void *)&A[i][0];
   void *Bcol = (void *)&B[0][k];
   int j;
   int result = 0;
   int NN = 4*n;
   for (j = 0; j < NN; j+=4){
        result += *((int *)(Arow+j)) * *((int *)Bcol);
        Bcol += NN;
   }
   return result;
}

$ gcc -O2 -S ics3.61.c -m32

.L4:
    movl    (%ebx), %ecx 
    addl    %esi, %ebx #esi->NN, ebx->Bcol
    imull   (%edi,%edx), %ecx #ecx=(*Bcol) * *(Arow+j), edi->Arow
    addl    $4, %edx #edx->j
    addl    %ecx, %eax #eax->result 
    cmpl    %edx, %esi 
    jg  .L4

3.62

A. 13
B. i: %edi; j: %ecx
C.

void transpose(int M, int A[M][M]){
    int i, j, *A1, t;
    for (i = 0; i < M; i++){
        A1 = &A[0][i];
        A_ = A[i];
        for (j = 0; j < i;){
            t = A_[j];
            A_[j] = *A1;
            j++;
            *A1 = t;
            A1 += M;
        }
    }
}

3.64

A.
8(%ebp)存放返回值的地址
12(%ebp)存放s1.a
16(%ebp)存放s1.p
B.

%ebp->
	s2.diff
	s2.sum
	s1.p
	s1.a
%esp->	&s2

C.
将结构的每一个成员逆序压入栈中
D.
预先从低到高地址分配结构各个成员的空间，在栈中retaddr+4的位置提供一个指向结构的指针作为返回地址

---

3.67

A.

e1.p: 0
e1.y: 4
e2.x: 0
e2.next: 4

B. 8个
C. up->next->y = *(up->next->p) - up->x;

3.68

#include <stdio.h>
#define MAX_INPUT 10
int good_echo()
{
    char s[MAX_INPUT];
    char *ret;
    do{
        ret = fgets(s, MAX_INPUT, stdin);
        if(ret == NULL){
            if(feof(stdin)) return 0;
            else return 1;
        }
        fputs(s, stdout);
    } while(true);
}
int main()
{
    return good_echo();
}

3.70

#define MIN_VAL -9223372036854775807-1
long traverse(tree_ptr tp)
{
    if (!tp)
        return MIN_VAL; //movabsq??
    long v = tp->val;
    long vLeft = traverse(tp->left);
    long vRight = traverse(tp->right);
    if (vLeft >= vRight) vRight = vLeft;
    if (vRight < v) vRight = v;
    return vRight;
}