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Download and try UltraCompare before you buy it! This download is the fully-functional version of the application for Windows and includes all features. Download UltraCompare v2024.1 (released 2025-02-10) | Hotfix information English Download 64-bit SHA256 355A5B4710721982BFF7A4C482C11624317CEA377166CA4F36778B146926F826 Deutsch German Download 64-bit SHA256 E64AD5F7FC7B5F8A1D81C126923DCE760B3A457AA59F9FAB26D3569D10EA81DC Italiano Italian Download 64-bit SHA256 59929236FE373436099DDD07474D30C85141CD8015C098F0046CB675324EC6B0 Español Spanish Download 64-bit SHA256 928324A1C9E08FE6878FC03D818462FF4B5DE8B5D49F886261710695B8B39401 français French Download 64-bit SHA256 F1341B92C3A75D78006B8E8666D4AD937CDF3218C7BE0957F18F4B39C2510115 Português brasileiro Brazilian Portuguese Download 64-bit SHA256 9D051109C90ADF08E71A8F0E304F5772E9C1F8F2FEDC6E5971822C9732BDCC26 日本 Japanese Download 64-bit SHA256 360D294D9BB560F1158A8EC8656E37119D57FF259D3D93818C1E1CFE22884349 한국어 Korean Download 64-bit SHA256 66FF6D64FE1D86610857C06F3F0ADC3DD176B015BC0D9F12F1C1F3BB800D6195 华语 Chinese Download 64-bit SHA256 577BC02A4F56FCB4269EC2FEF5428B7B85FCB91CDD1D815C01DE830093B669BE Download UltraCompare extension for Visual Studio 2022 Early access version (released 2024-11-15) English Download 64-bit SHA256 7E0C6A5B1A84E0777645938C24015A48B284F0DBFBBEBAD8D08AACD60199DF90 EXE or MSI?The EXE builds of our products are the preferred download for most of our users, running Windows 8.1 and later.We make MSI builds available for bulk or silent deployments (command line, GPO, SCCM, etc.). Get the UltraCompare MSI installer.Free trial periodThis application comes with a free, fully functional 30 day trial period. After the trial period expires, you must purchase a license to continue using it.

0x408000000xC08000000x834000000xC85800000Answer (Detailed Solution Below) Option 2 : 0xC0800000 Concept: In IEEE- 754 single precision format, a floating-point number is represented in 32 bits. Sign bit (MSB) Biased Exponent (E’) (8 bits) Normalized Mantissa (M’) (23 bits) Sign bit value 0 means positive number, and 1 means a negative number.The floating-point number can be obtained by formula: ± 1. M × 2(E-127)Data:Content of R1: 0x 42200000 (0x means Hexadecimal notation)Content of R2: 0x C1200000Calculation:Content of R1 in Hex (0x) is 42200000. After converting into binary, it can be represented in IEEE- 754 format as: 0 100 0010 0 010 0000 0000 0000 0000 0000 Sign bit is 0 i.e. the number is positiveBiased Exponent (E’) = 100 0010 0 = 132Normalized Mantissa (M) = 010 0000 0000 0000 0000 0000 = .25Therefore, the number in register R1 = + 1.25 * 2(132-127) = 1.25 × 32 = 40Content of R2 in Hex (0x) is C1200000. After converting into binary, it can be represented in IEEE- 754 format as: 1 100 0001 0 010 0000 0000 0000 0000 0000 Sign bit is 1 i.e. the number is negativeBiased Exponent (E’) = 100 0001 0 = 130Normalized Mantissa (M) = 010 0000 0000 0000 0000 0000 = .25Therefore, the number in register R1 = - 1.25 * 2(130-127) = -1.25 * 8 = -10R3 = R1/R2 = 40/-10 = -4Since the number is negative, Sign bit (MSB) = 1Converting 4 into binary of a floating point gives: (100.0)2Representing it into normalized form gives: (1.000000….) × 22Therefore, Mantissa is 23 bits of all 0sBiased Exponent (E’) = E+ 127 = 2+127 = 129 = (10000001)2It can be represented in IEEE- 754 format as: 1 100 0000 1 000 0000 0000 0000 0000 0000 Converting it into Hex format gives: 0x C0800000 The decimal floating-point number -40.1 represented using IEEE-754 32-bit representation and written in hexadecimal form is _____ 0xC22060000xC20066660xC20060000xC2206666Answer (Detailed Solution Below) Option 4 : 0xC2206666 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) In IEEE-754 format, 32-bit (single precision)(-1)s × 1.M × 2E – 127Calculation:Convert: 40.1 to binaryStep 1: convert 40 2 40 2 20 0 2 10 0 2 5 0 2 2 1 2 1 0 0 1 ↑ (40)10 = (101000)2Step 2: convert .1 to binary0.1 × 2 = 0.2 (0)0.2 × 2 = 0.4 (0)0.4 × 0.2 = 0.8 (0)0.8 × 0.2 = 1.6 (1)0.6 × 0.2 = 1.2 (1)0.2 × 0.2 = 0.4 (0) and so onGiven binary number is(40.1)10 = (101000.000110011001100…)2(40.1)10 = 1.0100 0000 1100 1100 … × 25Signed (1 bit) = 1 (given number is negative)Exponent (8 bit) = 5 + 127 =

× 10-1 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) Calculation:Given binary number is00111110011011010000000000000000Here, sign bit is 0. So, number is positive. 0 01111100 11011010000000000000000 Exponent bits = E = 01111100 = 124 (in decimal)Mantissa bits M = 11011010000000000000000In IEEE-754 format, 32-bit (single precision) (-1)s × 1.M × 2E – 127 = (-1)0 × 1.1101101 × 2124 – 127= 1.1101101 × 2-3= (1 + 2-1 + 2-2 + 2-4 + 2-5 + 2-7) × 2-3= 0.231 = 2.31 × 10-1 ≈ 2.27 × 10-1 In IEEE floating point representation, the hexadecimal number 0xC0000000 corresponds to –3.0–1.0–4.0–2.0Answer (Detailed Solution Below) Option 4 : –2.0 Concept:32-bit floating-point representation of a binary number in IEEE- 754 is Sign (1 bit) Exponent (8 bit) Mantissa bit (23 bits) Calculation:Binary number is0xC0000000 = (11000000000000000000000000000000)2Here, the sign bit is 1. So, the number is negative. 1 10000000 00000000000000000000000 Exponent bits = E = 10000000 = 128 (in decimal)Mantissa bits M = 00000000000000000000000In IEEE-754 format, 32-bit (single precision)(-1)s × 1.M × 2E – 127= (-1)1 × 1. 0 × 2128 – 127= -1 × 1.0 × 2= -2In IEEE floating-point representation, the hexadecimal number 0xC0000000 corresponds to -2. Let R1 and R2 be two 4-bit registers that store numbers in 2’s complement form. For the operation R1 + R2, which one of the following values of R1 and R2 gives an arithmetic overflow? R1 = 1011 and R2 = 1110R1 = 1100 and R2 = 1010R1 = 0011 and R2 = 0100R1 = 1001 and R2 = 1111Answer (Detailed Solution Below) Option 2 : R1 = 1100 and R2 = 1010 The correct answer is option 2.Concept:Stored numbers in registers R1 and R2 are in 2's complement form. Register size is 4 bits. The range of numbers in 2's complement form is -8 to +7. If R1 + R2, the result is out of the above range, then it is overflow.The given data,Given two four-bit registers R1 and R2.Option 1: R1 = 1011 and R2 = 1110False, R1 = 1 0 1 1 = -(0101)= -5+ R2 = 1 1 1 0 = -(0010)= -2----------------------------------------------- 1 0 0 1 = = -7 Here No overflow occurred, because sign bit is same for (R1 + R2 ).Option 2: R1 = 1100 and R2 = 1010True,R1 = 1 1 0 0 = -(0100)= -4+ R2 = 1 0 1 0 = -(0110)= -6 -------------------------------------------- 0 1 1 0 = = -10 Here Overflow occurred because the sign bit is different for (R1 + R2 ).Option 3: R1 = 0011 and R2 = 0100False,R1 = 0 0 1 1 = +(0011)= +3+ R2 = 0 1 0

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