machine using 12-bit two's complement arithmetic for signed integers short integers are encoded using 6 bits The following definitions are given: int x--307; unsigned u- x; short s (short) -101; short sx - (ahort)x; Fill in the empty boxes in the tables below. Hint: Be careful with the rules that Cuses for signed and unsigned ints. Show all your workd Decimal Binary Representation Expression Representation 0000 0000 0000 a) -307 b) c) d) u/ 2 e) > 1 g) (tu - 1)& u h) ((int) (u >> 11) +0) ((int) u) >> 11 Note: Write your final answers for corresponding empty cells like a.) a.i) b.i)

machine using 12-bit two's complement arithmetic for signed integers short integers are encoded using 6 bits The following definitions are given: int x--307; unsigned u- x; short s (short) -101; short sx - (ahort)x; Fill in the empty boxes in the tables below. Hint: Be careful with the rules that Cuses for signed and unsigned ints. Show all your workd Decimal Binary Representation Expression Representation 0000 0000 0000 a) -307 b) c) d) u/ 2 e) > 1 g) (tu - 1)& u h) ((int) (u >> 11) +0) ((int) u) >> 11 Note: Write your final answers for corresponding empty cells like a.) a.i) b.i)

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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1a.what is the output of these 2 programs #include <stdio.h>main() { int i=0; for( ;i<=2; ) printf(" %d",++i);} #include<stdio.h>main( ){ int i,j; for(i=0; i<2; i++){ for(j=0; j<2; j++){ printf("%d %d\n",i,j); break; } }} 1b. can someone show me how to do this problem the following two 6-bit binary values are two's complement signed integers. Represent the sum as a 6-bit two'scomplement value.001001111010
You are given two bit positions, I and j, along with the 32-bit numbers N and M. To insert Minto N, create a method that begins M at bit j and ends it at bit i. It is safe to presume that all of M can fit in the bits j through i. Therefore, you can assume that there are at least 5 bits between j and I if M = 10011. M could not completely fit between bit 3 and bit 2, so you would not, for instance, have j = 3 and I = 2.Input: N=10000000000, M=10011,i=2,j=6Output: N=10001001100
Suppose A = -14 and B = +6 (both in base 10) C=A*B What is the resulting product of your previous multiplication in FP decimal representation (simply use a decimal point)? Take the result you just produce and shift it by three digits to the right of the decimal point (e.g., if your number was 12.0 it now becomes 0.012). Now, express your shifted result as a single precision FP number using the IEEE 754 standard for single precision (of course you need to convert the number from decimal to binary). Please do not calculate more than the first 8 fraction bits.
THIS IS ALL FOR YHE SAME PROBLEM! For each of the operations below: - convert all values to the given number system - apply the operation within the given number system. (Just like the computer, you must return a value within the same number system.) - indicate whether an overflow has occurred i ) 74 + 12 in 8 digit unsigned binary ii 123 + 23 in 8 digit two’s complement binary iii −30 + 42 in 7 digit two’s complement binary iv 240 + 20 in 2 digit hexadecimal
1. What is the result when (7x + 2y) is multiplied to (7x – 2y)? A. 14x² - 4y² B. 14x² + 4y² C. 49x² - 4y² D. 49x² + 4y² 2. Which of the following expressions, when simplified, is equal to 1? A. 5²/5⁰ B. 0²⁵ C. (7x⁰)¹ D. (700x⁵)⁰
Let's Assume, X=224A0000 and Y=336B0000 are two Hex numbers. Using IEEE-754 floating point arithmetic perform the below calculations and give the answers in decimal: Y What is the Bias exponent of X? What is actual exponent of X? What is the Bias exponent of Y? What is actual exponent of y? Now, what would be the value of X+Y in decimal? Consider two decimal points of the final results after rounding-up. A sample answer: 4.23*10^3When you are typing the '*' symbol in the input field, it will appear as a '·' symbol.
This code is correct in reversing a 32-bit integer when looking at the binary form of the integer. But it uses 54 operations instead of 45. We are strictly limited to using only !, +, ^, &, <<, >>, |, ~. We cannot use constants that are anything other than 0x0 to 0xFF. We also cannot use loops or conditionals, in other words "straight line" code. I am unsure where I can simplify this code in order to get it into the required maximum number of operations (45). Will leave good review! int bitReverse(int x) {int result, mask1, mask2, mask4, mask8, mask16, min; min = 1 << 31;mask16 = ((0xff << 8) | 0xff);mask8 = (mask16 << 8) ^ mask16;mask4 = (mask8 << 4) ^ mask8;mask2 = (mask4 << 2) ^ mask4;mask1 = (mask2 << 1) ^ mask2;mask1 = ~mask1;mask2 = ~mask2;mask4 = ~mask4;mask8 = ~mask8;mask16 = ~mask16; result = ((x << 16) & mask16) | (((x & mask16) >> 16) & (~mask16));result = ((result << 8) & mask8) |…
4, in the fixed-point machine, the following statement is wrong ( ).A. In addition to the complement, the original code and the inverted code cannot indicate that the original code of −1 B. +0 is not equal to −0C. the inverse of +0 is not equal to the negative of −0 D. For the same machine word length, the complement code can represent a negative number more than the original code and the inverted code.
f(x)=x^2(x-4)(x+2)^2 Bounce(B) or Cross(C) (in ascending order of zeros): Incorrect (So type CBC or BBB or...)
Consider the following code which includes a parity check for the third digit: C=(000,011,101,110). You get the third digit by adding the first two and then using the remainder on division by 2. Can this code detect and count single errors? how can this be implemented?
To sign extend an unsigned integer from 8 bits to 16 bits, you would first fill the first 8 MSB bits with all ones, which would result in a total of 16 bits. This would bring the total number of bits to 16. Choose one from the following options: True or False?
Assume that D has the value 1010101010 and that the 5-bit generator G=10011 is used. What does R stand for? B. C. Does corruption exist? Publish your work. Repetition of parts a and b with the addition of the following: the numbers 1001010101, 0101101010, and 1010100000