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1 Use √A when expressing an address in a computer
. Unsigned number
B. Inverse code
C. Original code
D. Complementary code
2 The representation range and representation accuracy of floating-point numbers depend on
A. the encoding of the exponent and the encoding of the mantissa respectively
√B. the number of digits of the exponent and the number of digits of the mantissa
C. the length of the machine word and the number of digits of the exponent
D. the mantissa The number of digits and the number of exponents
3 Let G(x) = 1011, a certain (7,4) code is K1K2K3K4K5K6K7, when only K7 is wrong, the remainder obtained by CRC check is 001, when only K5 is wrong, the remainder obtained by CRC check is √A
. 100
B. 011
C. 010
D. 110
According to the cycle of the remainder of the CRC check, add a 0 after the remainder obtained when K7 makes an error, and then use the generator polynomial 1011 to do modulo 2 division to obtain the remainder when K6 makes an error.
0010/1011...010 Add a 0 after the obtained remainder, and then use the generator polynomial 1011 to do modulo 2 division to get the remainder 100/1011...100 when K5 makes an error. The answer is 100. Reference: https:
//
blog . csdn.net/qq_20115319/article/details/107310235
4 If a certain system 15*4=112 is established, the base system adopted by the system is
√A. 6
B. 9
C. 8
D. 7
Assuming that the p-base system is used, 15*4=112 is equivalent to:
(p + 5) * 4 = p^2 + p + 2
to solve p=-3 (round down) and p=6
Reference: https ://blog.csdn.net/xhyxxx/article/details/64132917
5 2^100 mod 7 =
A. 5
√B. 2
C. 4
D. 3
2＾3=8 mod 7 =1，
100 mod 3=1,
2＾100=2＊2＾99 mod 7=2.
6 The effective binary data to be encoded with a length of 5 bits is 11011, and the generator polynomial code used for CRC encoding is 1011, and the obtained CRC encoding is A. 11011010 B. 11011101
C.
11011100
√D
. 11011001
Divide 11011000 modulo 2 by 1011 to get the remainder 001.
1101 /1011=0110, one bit to the left
110 1 /1011=0110, one bit to the left
110 0 /1011=0111, one bit to the left
111 0 /1011=0101, one bit to the left
101 0 /1011=0 001 . The remainder has fewer digits than the divisor and is the final remainder.
7 The highest 8 bits of a certain hexadecimal floating-point number A3D00000 are exponent codes (including 1-digit character), and the mantissa is the lowest 24 bits (including 1-digit character). The decimal value of the point is
A. -0.375×2^(-35)
B. -0.625×2^(-93)
√C. -0.375×2^(-93)
D. 0.625×2^(-35)
It should be noted that this is not IEEE754, do not use -127 and do not
add 1 of the highest digit in the mantissa
. +0.125=0.375. The sign bit is all negative.
Answer: (-1)＊2＾(-93)＊(-0.375)
Reference: https://blog.csdn.net/qq_20115319/article/details/107310235
8 The address numbers in the memory are respectively 1000#, 1001#, 1002#, and 1003 in the 4 consecutive storage units, and the byte data stored respectively are 1A, 2B, 3C, 4D. If the data word length is 32 bits, the memory adopts is the little-endian alignment mode, the data values stored in these 4 storage units should be parsed as
√A. 4D3C2B1A
B. D4C3B2A1
C. 1A2B2C3D
D. A1B2C3D4
9 A two-complement integer with a word length of 8 bits is 11011010, then the standard frame shift of the number is
A. 00111010
B. 11011010
C. 10111010
√D. 01011010
The frame shift is used to represent the exponent code of the floating-point number, and the integer representation. The relationship between the code shift and the complement code is: the sign bit of the n-bit integer complement code is inverted to obtain the corresponding code shift. Regarding the principle, I made a schematic diagram as follows:
10 Two complements 0A2B and E16A with a word length of 16 bits, and the results after sign extension to 32 bits are
A. 00000A2B and 0000E16A
√B. 00000A2B and FFFFE16A
C. FFFF0A2B and 0000E16A
D. 11110A2B and FFFFE16A
11 For floating-point numbers in IEEE754 format, the following description is correct:
A. The exponent code is represented by shift code, and the mantissa is represented by complement code
√B. The exponent code is represented by shift code, and the mantissa is represented by original code
C. Both the exponent code and the mantissa are represented by The original code represents
D. Both the exponent code and the mantissa are represented by complement code
The order code is represented by a frame shift, and the mantissa is represented by an original code.
Supplement : The single-precision floating-point number is 32 bits long, the sign bit S occupies 1 bit, the exponent E occupies 8 bits, the mantissa M occupies 23 bits, and the exponent offset is 127, expressed by shifting code; the double-precision floating-point number is 64 bits long , The sign bit S occupies 1 bit, the exponent E occupies 11 bits, the mantissa M occupies 52 bits, and the exponent offset is 1023, which is expressed by shifting codes; it is agreed that
one bit 1 is implied to the left of the decimal point of M, and the complete mantissa form is 1.M. However, only M is saved when performing floating-point data representation. Therefore, the length of the above-mentioned single-precision mantissa is actually 24, and the length of the double-precision mantissa is actually 53, which is why the offset is not 128 or 1024;
12 For the binary code 10001101 with a word length of 8 bits, the following statements are wrong:
A. If the code is an unsigned number, its decimal value is +141
√B. If the code is a standard shift number, its decimal value is is +115
C. If the code is an original number, its true decimal value is -13
D. If the code is a complement number, its true decimal value is -115
B. +13。
13 In the following description of Hamming check, the correct description is
√A. The length of the data to be encoded is 4 bits, and adding 3 check codes can detect and correct the error of only 1 bit of data. B.
Haiming It is clear that the code distance of the check code has nothing to do with the number of check digits
C. When divided into r groups, r bits can be obtained to indicate error codes, so it is possible to distinguish 2 r cases where only 1 bit is wrong.
D. Under any conditions , the Hamming check cannot detect the simultaneous error of multiple bits of data
B. In general, increasing the checksum also increases the code distance.
C. Remove the bit occupied by the check code. When divided into r groups, r bits can be obtained to indicate error codes, so (2＾rr) cases where only 1 bit is wrong can be distinguished.
D. Divide the effective information into several groups according to certain rules, arrange a parity bit for each group, and perform a parity test to provide multi-bit error detection information to point out which bit is most likely to be wrong, so as to correct it. In essence, Hamming check is a kind of multiple check.
14 If the mantissa of the floating-point number is represented by 5-bit complement code (where the sign bit is 1 bit), the normalized mantissa in the following mantissas is √A. 10000
and 01001
B. 01011 and 11010
C. 01100 and 11110
D. 11011 and 01011
A. Original code: 1.0000, 0.1001, all normalized.
15 In the following descriptions about the relationship between complement code and code shift, the error is A. Complement code and code shift with the same number of digits
have the same data representation range.
Indicates a fixed-point number
C. Complement code and code shift of the same number have the same value part, but the sign is opposite
√D. The complement code and code shift of zero are the same
D. Different, the frame shift of zero is 1.0000, and the sign bit must be reversed.
multiple choice
1 Known fixed-point decimal [X] complement = 1.X1X2X3, if X≤-0.75 is required, then the following requirements are
√A. X1X2X3 = 000
√B. X1X2X3 = 001
C. X1X2X3 =101
√D. X1X2X3=010
Just convert it to a decimal.
2 Suppose the exponent code of a normalized floating-point number is k+1 bits (including a sign bit), and the mantissa is n+1 bits (including a sign bit). If both the exponent code and the mantissa are represented by complement code data, the following about In the description of the floating-point number range, the correct one is
√A. The smallest positive number that can be expressed is
√B. The largest negative number that can be expressed is
√C. The largest positive number that can be expressed is
√D. The smallest negative number that can be expressed is
3 What does not appear in the IEEE 754 floating-point machine number is
√A. The highest bit in the mantissa 1
B. The sign bit of the mantissa
C. The sign bit of the exponent code
√D. The base
C. The frame shift is also signed.
4 In the following description about parity check, the error is
A. Only 1 parity bit is required
B. The error conclusion obtained by the detection is reliable
√C. 1 bit error can be corrected
√D. The error-free conclusion obtained by the detection is reliable
5 The word length of the computer is 16 bits, and the integer is represented by complement code. The correct one in the following description about the data range is
√A. The smallest positive number that can be represented is 1
√B. The largest negative number that can be represented is -1
√C. The largest positive number that can be represented is (2^15) - 1
√D. The smallest negative number that can be represented is - (2 ^15)
fill in the blank
1 The machine number in 32-bit IEEE754 format corresponding to the decimal number 5 is (40A00000) H (expressed in hexadecimal, you do not need to write the last H when filling in the answer)
2 All codewords of an error correcting code are 0000000000, 0000011111, 1111100000, 1111111111. The code distance of this code is (5)
3. The valid binary data to be encoded with a word length of 5 bits is 10011. The generator polynomial code used for CRC encoding is 1011, and the obtained encoding result is (10011100) (fill in the binary form)
4 When representing a decimal number -129 as a complement, how many binary digits should be used at least (9)
To be added.