Fundamentals of Information Systems (Looseleaf)
9th Edition
ISBN: 9781337099042
Author: STAIR
Publisher: Cengage
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Chapter 7, Problem 14SAT
Program Plan Intro
To choose the correct option to fill in the blank.
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1 typedef struct node* {
2
struct node* next;
3
char* key;
4
char* val;
5} node_t;
6
7 char* find_node (node_t* node, char* key_to_find) {
while(strcmp (node->key, key_to_find ) != 0 ) {
node = node->next;
8
9
10
}
11
return node->val;
12 }
Match each of the assembler routines on the left with the equivalent C function on
the right. Write the name of the label (e.g., foo) to the right of the corresponding
function. Note: shrq is the logical right shift instruction, and sarq is the arithmetic
right shift instruction.
foo1:
leaq
0(,%rdi, 8), %rax
long choice1 (long x)
{
ret
return x -
8 >8;
foo3:
}
movq
sarq
%rdi, %rax
$8, %rax
long choice4 (long x)
ret
{
return x*256;
}
foo4:
long choice5 (long x)
leaq
-8 (%rdi), %rax
{
ret
return x-8;
}
long choice6 (long x)
foo5:
{
leaq
-8 (%rdi), %rax
return x+8;
shrq
$63, %rax
}
ret
Given the variables and code in the text below, identify where in memory they will
live once the code is compiled.
1 char
big_array [1L<<24]; /* 16 MB */
2 GB *
:/
2 char huge_array [1L<<31]; /*
3
4 int global = 0;
5
6 int useless () { return 0; }
7
8 int main()
9 {
10
void *p1, p2, *p3, *p4;
int local =
0;
malloc (1L << 28); /* 256 MB *,
11
12
p1
13
p2
=
malloc (1L << 8);
/* 256 B *
14
p3
15
p4
=
malloc (1L << 32);
malloc (1L << 8);
/* 4
GB *
*/
/* 256
B */
16 }
Note: *pN is the thing at which pN points.
1. big_array
2. huge_array
3. global
4. useless
5. void* p1
6. *p1
7. void* p2
8. *p2
9. void* p3
10. *p3
11. void* p4
12. *p4
Chapter 7 Solutions
Fundamentals of Information Systems (Looseleaf)
Ch. 7 - Prob. 1LOCh. 7 - Prob. 2LOCh. 7 - Prob. 3LOCh. 7 - Prob. 4LOCh. 7 - Prob. 5LOCh. 7 - Prob. 6LOCh. 7 - Prob. 7LOCh. 7 - Prob. 8LOCh. 7 - Prob. 9LOCh. 7 - Prob. 1.1RQ
Ch. 7 - Prob. 1.2RQCh. 7 - Prob. 1.1CTQCh. 7 - Prob. 1.2CTQCh. 7 - Prob. 2.1RQCh. 7 - Prob. 2.2RQCh. 7 - Prob. 2.1CTQCh. 7 - Prob. 2.2CTQCh. 7 - Prob. 3.1RQCh. 7 - Prob. 3.2RQCh. 7 - Prob. 3.1CTQCh. 7 - Prob. 3.2CTQCh. 7 - What sort of training and experience is needed by...Ch. 7 - Prob. 4.2RQCh. 7 - Prob. 4.1CTQCh. 7 - Prob. 4.2CTQCh. 7 - Prob. 1SATCh. 7 - Prob. 2SATCh. 7 - Prob. 3SATCh. 7 - Prob. 4SATCh. 7 - Prob. 5SATCh. 7 - Prob. 6SATCh. 7 - Prob. 7SATCh. 7 - Prob. 8SATCh. 7 - Prob. 9SATCh. 7 - Prob. 10SATCh. 7 - Prob. 11SATCh. 7 - Prob. 12SATCh. 7 - Prob. 13SATCh. 7 - Prob. 14SATCh. 7 - Prob. 15SATCh. 7 - Prob. 16SATCh. 7 - Prob. 17SATCh. 7 - Prob. 1RQCh. 7 - Prob. 2RQCh. 7 - Prob. 3RQCh. 7 - Prob. 4RQCh. 7 - Prob. 5RQCh. 7 - Prob. 6RQCh. 7 - Prob. 7RQCh. 7 - Prob. 8RQCh. 7 - Prob. 9RQCh. 7 - Prob. 10RQCh. 7 - Prob. 11RQCh. 7 - Prob. 12RQCh. 7 - Prob. 13RQCh. 7 - Prob. 14RQCh. 7 - Prob. 15RQCh. 7 - Prob. 16RQCh. 7 - Prob. 17RQCh. 7 - Prob. 18RQCh. 7 - Prob. 19RQCh. 7 - Prob. 20RQCh. 7 - Prob. 21RQCh. 7 - Prob. 22RQCh. 7 - Prob. 23RQCh. 7 - Prob. 1DQCh. 7 - Prob. 2DQCh. 7 - Prob. 3DQCh. 7 - Prob. 4DQCh. 7 - Prob. 5DQCh. 7 - Prob. 6DQCh. 7 - Prob. 7DQCh. 7 - Prob. 8DQCh. 7 - Prob. 9DQCh. 7 - Prob. 10DQCh. 7 - Prob. 11DQCh. 7 - Prob. 12DQCh. 7 - Prob. 13DQCh. 7 - Prob. 1PSECh. 7 - Prob. 2PSECh. 7 - Prob. 3PSECh. 7 - Prob. 1WECh. 7 - Prob. 2WECh. 7 - Prob. 3WECh. 7 - Prob. 1CECh. 7 - Prob. 2CECh. 7 - Prob. 3CECh. 7 - Prob. 1.1CSCh. 7 - Prob. 1.2CSCh. 7 - Prob. 1.3CSCh. 7 - Prob. 1.4CSCh. 7 - Prob. 1.5CSCh. 7 - Prob. 2.1CSCh. 7 - Prob. 2.2CSCh. 7 - Prob. 2.3CS
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- The next problem concerns the following C code: /copy input string x to buf */ void foo (char *x) { char buf [8]; strcpy((char *) buf, x); } void callfoo() { } foo("ZYXWVUTSRQPONMLKJIHGFEDCBA"); Here is the corresponding machine code on a Linux/x86 machine: 0000000000400530 : 400530: 48 83 ec 18 sub $0x18,%rsp 400534: 48 89 fe mov %rdi, %rsi 400537: 48 89 e7 mov %rsp,%rdi 40053a: e8 di fe ff ff callq 400410 40053f: 48 83 c4 18 add $0x18,%rsp 400543: c3 retq 400544: 0000000000400544 : 48 83 ec 08 sub $0x8,%rsp 400548: bf 00 06 40 00 mov $0x400600,%edi 40054d: e8 de ff ff ff callq 400530 400552: 48 83 c4 08 add $0x8,%rsp 400556: c3 This problem tests your understanding of the program stack. Here are some notes to help you work the problem: ⚫ strcpy(char *dst, char *src) copies the string at address src (including the terminating '\0' character) to address dst. It does not check the size of the destination buffer. • You will need to know the hex values of the following characters:arrow_forwardConsider the following assembly code for a C for loop: movl $0, %eax jmp .L2 .L3: addq $1, %rdi addq %rsi, %rax subq $1, %rsi .L2: cmpq %rsi, %rdi jl .L3 addq ret %rdi, %rax Based on the assembly code above, fill in the blanks below in its corresponding C source code. Recall that registers %rdi and %rsi contain the first and second, respectively, argument of a function. (Note: you may only use the symbolic variables x, y, and result in your expressions below do not use register names.) long loop (long x, long y) { long result; } for ( } return result; __; y--) {arrow_forwardIn each of the following C code snippets, there are issues that can prevent the compiler from applying certain optimizations. For each snippet: Circle the line number that contains compiler optimization blocker. ⚫ Select the best modification to improve optimization. 1. Which line prevents compiler optimization? Circle one: 2 3 4 5 6 Suggested solution: ⚫ Remove printf or move it outside the loop. Remove the loop. • Replace arr[i] with a constant value. 1 int sum (int *arr, int n) { 2 int s = 0; 3 for (int i = 0; i < n; i++) { 4 5 6 } 7 8 } s = arr[i]; printf("%d\n", s); return s; 234206 2. Which line prevents compiler optimization? Circle one: 2 3 4 5 6 Suggested solution: Move or eliminate do_extra_work() if it's not necessary inside the loop. Remove the loop (but what about scaling?). ⚫ Replace arr[i] *= factor; with arr[i] = 0; (why would that help?). 1 void scale (int *arr, int n, int factor) { 5 6 } for (int i = 0; i < n; i++) { rr[i] = factor; do_extra_work ();arrow_forward
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