Suppose 0002 is the base address of a segment. Now consider two memory locations are 0002F and 1002F. Which statement is true? 0002F is inside the segment 1002F is inside the segment 0002F is inside but 1002F is outside the segment. None of the above

Suppose 0002 is the base address of a segment. Now consider two memory locations are 0002F and 1002F. Which statement is true? 0002F is inside the segment 1002F is inside the segment 0002F is inside but 1002F is outside the segment. None of the above

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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2. Write a simple encryption/decryption program. Function encrypt takes a character pointer as a parameter and uses pointer-subscript notation to change the value in the address pointed to by adding 1 to it. Function decrypt takes a character pointer as a parameter and uses pointer notation to change the value in the address pointed to by subtracting 1 from it. Function main calls functions encrypt and decrypt and prints the encrypted string. // A simple encryption/decryption program #include "stdafx.h" #include using namespace std; // write the prototype for function encrypt // write the prototype function decrypt int main() { // create a string to encrypt char string[] = "this is a secret!"; cout « "Original string is : " « string « endl; encrypt(string); cout « "Encrypted string is : « string <« endl; decrypt(string); cout <« "Decrypted string is : «string<< endl;
Write out how to solve the following code using any addressing mode for (3,3) GPR. a[] is address0x100, y is address 0x300, z is address 0x400, and x is using r0. Write comments besides each line ofcode. Short a[50];Int y,z;for( x=0; x<20; x++){y = a[x + 3];z = y + a[x+10];}
The program below uses pointer arithmetic to determine the size of a 'char'variable. By using pointer arithmetic we can find out the value of 'cp' and thevalue of 'cp+1'. Since cp is a pointer, this addition involves pointer arithmetic:adding one to a pointer makes the pointer point to the next element of the sametype.For a pointer to a char, adding 1 really just means adding 1 to the address, butthis is only because each char is 1 byte.1. Compile and run the program and see what it does.2. Write some code that does pointer arithmetic with a pointer to an int anddetermine how big an int is.3. Same idea – figure out how big a double is, by using pointer arithmetic andprinting out the value of the pointer before and after adding 1.4. What should happen if you added 2 to the pointers from exercises 1through 3, instead of 1? Use your program to verify your answer.
You are given the following function moveIndexes, where NR and NC are numerical constants. The function receives as parameters rp and cp, two pointers to int. The goal of the function is to modify the values pointed by rp and cp as follows: if rp and cp originally point to the indexes of a certain cell of the matrix, after function invocation they must point to the indexes of a cell at a distance delta from the original one. int moveIndexes (int M[NR] [NC], int *rp, int *ocp, int delta) ( int i, j, *p, *p0; i = *rp; j = *cp; if (i=NR |l j=NC) return l; // error - nothing done p0 = & (M[0] [0]); p = & (M[i][j]); p = p+delta; if (p & (M[NR-1] [NC-1])) return 1; // error nothing done *rp = (p - p0) / NC; *cp = (p - p0) & NC; return 0;
Write a short C program that declares and initializes (to any value you like) a double, an int, and a char. Next declare and initialize a pointer to each of the three variables. Your program should then print the address of, and value stored in, and the memory size (in bytes) of each of the six variables.
Find the PA of the memory location and its contents after the execution of thefollowing assuming that DS=1512h. MOV AL, 99h MOV [3518], AL
6. Write down the three-address codes for the following code segment. S := 0, I := 0 WHILE I < 10 DO S := s + X[I] I := I + 1 ENDWHILE
You are writing code that is equivalent to: X = A / (5 + B) Assume signed bytes and the following first 3 lines of code: mov al,A mov bl,B add bl,5 What is the one line of code that will perform the division correctly?
Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at? my answer: (correct) the second is equal to = 8 the third is equal to = 16 help me find the last word
2) Write a program that finds the position of the first 1 in an 8-bit data item. The data is scanned from D0 to D7. Write the result into 0x30. For example, if the data is equal to 11110000, then the program should write 0x04 into address 0x30.
write a program to count the number of data elements equal (FF)H in a set of data starting at address (0100)H and ended with the value of (00)H.
Question 10 Match the data types on the left with range of values they can represent. The range depends on the number of bytes (8-bits per byte). Assume that a modern 64-bit operating system is being used. NOTE: You can verify the size of each data type using the sizeof operator. unsigned long long float double address of char address of long long address of double -9,223,372,036,854,775,808 [Choose] -32,768 to 32,767 0 to 65,535 0 to 4,294,967,295 -128 to 127 (+/-)1.2E-38 to (+/-)3.4E+38 0 to 18,446,744,073,709,551,615 (+/-)2.3E-308 to (+/-)1.7E+308 -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 -2,147,483,648 to 2,147,483,647 0 to 255 [Choose] [Choose]