Introduction To Algorithms, Third Edition (international Edition)
Introduction To Algorithms, Third Edition (international Edition)
3rd Edition
ISBN: 9780262533058
Author: Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein
Publisher: TRILITERAL
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Chapter B.2, Problem 2E
Program Plan Intro

To prove that the relation, ab(modn) is aequivalent.

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Character Hex value | Character Hex value Character Hex value 'A' 0x41 'J' Ox4a 'S' 0x53 'B' 0x42 'K' 0x4b "T" 0x54 0x43 'L' Ox4c 'U' 0x55 0x44 'M' 0x4d 'V' 0x56 0x45 'N' Ox4e 'W' 0x57 0x46 '0' Ox4f 'X' 0x58 0x47 'P' 0x50 'Y' 0x59 0x48 'Q' 0x51 'Z' Ox5a 'T' 0x49 'R' 0x52 '\0' 0x00 Now consider what happens on a Linux/x86 machine when callfoo calls foo with the input string "ZYXWVUTSRQPONMLKJIHGFEDCBA". A. On the left draw the state of the stack just before the execution of the instruction at address Ox40053a; make sure to show the frames for callfoo and foo and the exact return address, in Hex at the bottom of the callfoo frame. Then, on the right, draw the state of the stack just after the instruction got executed; make sure to show where the string "ZYXWVUTSRQPONMLKJIHGFEDCBA" is placed and what part, if any, of the above return address has been overwritten. B. Immediately after the ret instruction at address 0x400543 executes, what is the value of the program counter register %rip?…
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
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