NEW PERSP.ON COMP.CONCEPTS-PACKAGE
17th Edition
ISBN: 9781337761116
Author: Carey
Publisher: CENGAGE L
expand_more
expand_more
format_list_bulleted
Question
Chapter 3, Problem 4CP4
Program Plan Intro
To create style rules in file pcg_paper.css and create a layout for PCG sample page.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
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
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:
Chapter 3 Solutions
NEW PERSP.ON COMP.CONCEPTS-PACKAGE
Ch. 3.1 - Prob. 1QCCh. 3.1 - Prob. 2QCCh. 3.1 - Prob. 3QCCh. 3.1 - Prob. 4QCCh. 3.1 - Prob. 5QCCh. 3.1 - Prob. 6QCCh. 3.1 - Prob. 7QCCh. 3.1 - Prob. 8QCCh. 3.1 - Prob. 9QCCh. 3.2 - Prob. 1QC
Ch. 3.2 - Prob. 2QCCh. 3.2 - Prob. 3QCCh. 3.2 - Prob. 4QCCh. 3.2 - Prob. 5QCCh. 3.2 - Prob. 6QCCh. 3.2 - Prob. 7QCCh. 3.2 - Prob. 8QCCh. 3.2 - Prob. 9QCCh. 3.3 - Prob. 1QCCh. 3.3 - Prob. 2QCCh. 3.3 - Prob. 3QCCh. 3.3 - Prob. 4QCCh. 3.3 - Prob. 5QCCh. 3.3 - Prob. 6QCCh. 3.3 - Prob. 7QCCh. 3 - Prob. 1RACh. 3 - Prob. 2RACh. 3 - Prob. 3RACh. 3 - Prob. 4RACh. 3 - Prob. 5RACh. 3 - Prob. 6RACh. 3 - Prob. 7RACh. 3 - Prob. 8RACh. 3 - Prob. 9RACh. 3 - Prob. 10RACh. 3 - Prob. 11RACh. 3 - Prob. 12RACh. 3 - Prob. 13RACh. 3 - Prob. 14RACh. 3 - Prob. 15RACh. 3 - Prob. 16RACh. 3 - Prob. 1CP1Ch. 3 - Prob. 2CP1Ch. 3 - Prob. 3CP1Ch. 3 - Prob. 4CP1Ch. 3 - Prob. 5CP1Ch. 3 - Prob. 6CP1Ch. 3 - Prob. 7CP1Ch. 3 - Prob. 8CP1Ch. 3 - Prob. 9CP1Ch. 3 - Prob. 10CP1Ch. 3 - Prob. 11CP1Ch. 3 - Prob. 12CP1Ch. 3 - Prob. 13CP1Ch. 3 - Prob. 14CP1Ch. 3 - Prob. 15CP1Ch. 3 - Prob. 16CP1Ch. 3 - Prob. 17CP1Ch. 3 - Prob. 18CP1Ch. 3 - Prob. 19CP1Ch. 3 - Prob. 20CP1Ch. 3 - Prob. 1CP2Ch. 3 - Prob. 2CP2Ch. 3 - Prob. 3CP2Ch. 3 - Prob. 4CP2Ch. 3 - Prob. 5CP2Ch. 3 - Prob. 6CP2Ch. 3 - Prob. 7CP2Ch. 3 - Prob. 8CP2Ch. 3 - Prob. 9CP2Ch. 3 - Prob. 10CP2Ch. 3 - Prob. 11CP2Ch. 3 - Prob. 12CP2Ch. 3 - Prob. 13CP2Ch. 3 - Prob. 14CP2Ch. 3 - Prob. 15CP2Ch. 3 - Prob. 16CP2Ch. 3 - Prob. 17CP2Ch. 3 - Prob. 18CP2Ch. 3 - Prob. 19CP2Ch. 3 - Prob. 20CP2Ch. 3 - Prob. 1CP3Ch. 3 - Prob. 2CP3Ch. 3 - Prob. 3CP3Ch. 3 - Prob. 4CP3Ch. 3 - Prob. 5CP3Ch. 3 - Prob. 6CP3Ch. 3 - Prob. 7CP3Ch. 3 - Prob. 8CP3Ch. 3 - Prob. 9CP3Ch. 3 - Prob. 10CP3Ch. 3 - Prob. 11CP3Ch. 3 - Prob. 12CP3Ch. 3 - Prob. 13CP3Ch. 3 - Prob. 14CP3Ch. 3 - Prob. 1CP4Ch. 3 - Prob. 2CP4Ch. 3 - Prob. 3CP4Ch. 3 - Prob. 4CP4Ch. 3 - Prob. 5CP4Ch. 3 - Prob. 6CP4
Knowledge Booster
Similar questions
- Consider 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_forward123456 A ROP (Return-Oriented Programming) attack can be used to execute arbitrary instructions by chaining together small pieces of code called "gadgets." Your goal is to create a stack layout for a ROP attack that calls a function located at '0x4018bd3'. Below is the assembly code for the function 'getbuf', which allocates 8 bytes of stack space for a 'char' array. This array is then passed to the 'gets' function. Additionally, you are provided with five useful gadgets and their addresses. Use these gadgets to construct the stack layout. Assembly for getbuf 1 getbuf: sub mov $8, %rsp %rsp, %rdi call gets add $8, %rsp ret #Allocate 8 bytes for buffer #Load buffer address into %rdi #Call gets with buffer #Restore the stack pointer #Return to caller Stack Layout each 8-byte (fill in section) Address Value (8 bytes) 0x7fffffffdfc0 0x7fffffffdfb8 0x7fffffffdfb0 0x7fffffffdfa8 0x7fffffffdfa0 0x7fffffffdf98 0x7fffffffdf90 0x7fffffffdf88 Gadgets Address Gadget Ox4006a7 pop %rdi; ret Ox4006a9…arrow_forward
- Problem 1 [15 points] The code below is buggy. Assume the code compiles. Briefly: 1). Identify the problem with the code (e.g., can access memory out of bounds) and 2). Suggest a solution (e.g., check the length). Question 1 1 #define BLENGTH 5 2 int b[BLENGTH]; 3 void copy_from_global_int_array_b (int n, int* dest) { 4 5 } *dest = b[n]; ==arrow_forwardWhich statement regarding SGA_MAX_SIZE is true? SGA_MAX_SIZE is modifiable after an instance is started, only when Automatic Memory Management is used. SGA_MAX_SIZE is not dyamically modifiable. SGA_MAX_SIZE is ignored when MEMORY_TARGET > 0. SGA-MAX_SIZE must be specified when SGA_TARGET > 0arrow_forwardExplian this C program #include <stdio.h> unsigned int rotateRight(unsigned int num, unsigned int bits) { unsignedint bit_count =sizeof(unsignedint) *8; bits = bits % bit_count; // Handle cases where bits >= bit_count return (num >> bits) | (num << (bit_count - bits)); } int main() { unsignedint num, bits; printf("Enter a number: "); scanf("%u", &num); printf("Enter the number of bits to shift: "); scanf("%u", &bits); printf("After rotation: %u\n", rotateRight(num, bits)); return0; }arrow_forward
- Explian thiS C program #include<stdio.h> int countSetBits(int n) { int count = 0; while (n) { count += n & 1; n >>= 1; } return count;} int main() { int num; printf("Enter a number: "); scanf("%d", &num); printf("Output: %d units\n", countSetBits(num)); return 0;}arrow_forwardPlease provide the Mathematica codearrow_forwardExplian this C program code. #include <stdio.h> void binary(unsigned int n) { if (n /2!=0) { binary(n /2); } printf("%d", n %2); } int main() { unsignedint number =33777; unsignedchar character ='X'; printf("Number: %u\n", number); printf("Binary: "); binary(number); printf("\nDecimal: %u\nHexadecimal: 0x%X\n\n", number, number); printf("Character: %c\n", character); printf("ASCII Binary: "); binary(character); printf("\nASCII Decimal: %u\nASCII Hexadecimal: 0x%X\n", character, character); return0; }arrow_forward
- Design a dynamic programming algorithm for the Longest Alternating Subsequence problem described below: Input: A sequence of n integers Output: The length of the longest subsequence where the numbers alternate between being larger and smaller than their predecessor The algorithm must take O(n²) time. You must also write and explain the recurrence. Example 1: Input: [3, 5, 4, 1, 3, 6, 5, 7, 3, 4] Output: 8 ([3, 5, 4, 6, 5, 7, 3, 4]) Example 2: Input: [4,7,2,5,8, 3, 8, 0, 4, 7, 8] Output: 8 ([4, 7, 2, 5, 3, 8, 0,4]) (Take your time with this for the subproblem for this one)arrow_forwardDesign a dynamic programming algorithm for the Coin-change problem described below: Input: An amount of money C and a set of n possible coin values with an unlimited supply of each kind of coin. Output: The smallest number of coins that add up to C exactly, or output that no such set exists. The algorithm must take O(n C) time. You must also write and explain the recurrence. Example 1: Input: C24, Coin values = = [1, 5, 10, 25, 50] Output: 6 (since 24 = 10+ 10+1+1 +1 + 1) Example 2: Input: C = 86, Coin values = [1, 5, 6, 23, 35, 46, 50] Output: 2 (since 86 = 46+35+5)arrow_forwardDesign a dynamic programming algorithm for the Longest Common Subsequence problem de- scribed below Input: Two strings x = x1x2 xm and y = Y1Y2... Yn Output: The length of the longest subsequence that is common to both x and y. . The algorithm must take O(m n) time. You must also write and explain the recurrence. (I want the largest k such that there are 1 ≤ i₁ < ... < ik ≤ m and 1 ≤ j₁ < ... < jk ≤ n such that Xi₁ Xi2 Xik = Yj1Yj2 ··· Yjk) Example 1: Input: x = 'abcdefghijklmnopqrst' and y = 'ygrhnodsh ftw' Output: 6 ('ghnost' is the longest common subsequence to both strings) Example 2: Input: x = 'ahshku' and y = ‘asu' Output: 3 ('asu' is the longest common subsequence to both strings)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
New Perspectives on HTML5, CSS3, and JavaScriptComputer ScienceISBN:9781305503922Author:Patrick M. CareyPublisher:Cengage Learning
New Perspectives on HTML5, CSS3, and JavaScript
Computer Science
ISBN:9781305503922
Author:Patrick M. Carey
Publisher:Cengage Learning