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# Required questions for group planning (machine level representation of data) (detailed analysis is attached in the update)

2022-05-15 07:47:48【Cold foam】

## One 、 Single topic selection

**1. In the following code , The only representation of zero is （C）**

A. Inverse code

B. Original code

C. Complement code

D. Source code and shift code

analysis ： This question belongs to the concept question , Four coding methods of computer numerical data are investigated . The four common codes of a computer are the original code 、 Inverse code 、 Complement and shift , The complement and shift codes are unique to the representation of zero , In this question, only C Options .

**2. Suppose the true value of a number is -100 1010B, Inside the computer, it is expressed as 1011 0110B, The number is encoded as （D）**

A. Shift the code

B. Original code

C. Inverse code

D. Complement code

analysis ： This question belongs to the concept question , Four coding methods of computer numerical data are investigated .-100 1010B The original code of is expressed as 1100 1010B, The inverse code is expressed as 1011 0101B, The complement is expressed as 1011 0110B, Code shift is expressed as 0111 0110B, Therefore, it can be seen that the code given in the question is complement .

**3. Consider the following code ： short si=-8196; unsigned short usi=si; If the 2 Byte representation short Type variable , After executing the above statement usi The value of is （C）**

A.8196

B.34572

C.57340

D.57339

analysis ： This question belongs to comprehension , Investigate the conversion between unsigned integer and integer .Si=-8196, Then use 16 Bit binary source code is expressed as 1010 0000 0000 0100B, Converted to a complement in the computer, expressed as 1101 1111 11111100B, In the second sentence usi The definition of is equivalent to si The number of machines has been reinterpreted , Interpreted as an unsigned short integer , Therefore, the corresponding true value is 2pow(16)-2pow(13)-2pow(1)-2pow(0)-1=65536-8192-2-1-1=57340.

**4. About IEEE754 Description of single precision , The wrong is （B）**

A. The order code adopts code shift

B. The mantissa is a complement

C. It is convenient for software transplantation

D. The significant number of mantissa is 24 position

analysis ： This question belongs to the concept question , Investigate IEEE754 Related concepts of single precision number .IEEE754 The standard specifies that the single precision number is 32 Bit representation , among 1 Bit is sign bit ,8 Bit order code ,23 Bits are mantissa . The order code is represented by binary shift code, which is convenient for order comparison , Mantissa is expressed in binary original code and pure decimal . This question is easy to make mistakes D term , however IEEE754 The mantissa contains a hidden bit , So with 23 Significant bits can represent 24 Significant digits .

**5.float type （ namely IEEE754 Single-precision floating-point ） The maximum positive number that can be expressed is （C）**

A.2pow(126)-2pow(103)

B.2pow(127)-2pow(104)

C.2pow(127)-2pow(103)

D.2pow(128)-2pow(104)

analysis ：IEEE754 The significant numbers of mantissa in the standard are 24 position , The significant digits of the order code are 8 position . Indicates the maximum positive number , Symbol bit 1; Take the mantissa, everyone 1, by 2pow(-1)+2pow(-2)+…2pow(-24)=1-2pow(-24), Take all the order codes 1, by 2pow(7)-1=127, So the end result is (1-2pow(-24))*2pow(127)=2pow(127)-2pow(103).

**6. In the storage and retrieval of Chinese characters , Need to adopt （C）**

A. input code

B. Location code

C. Internal code

D. GB code

analysis ： This question belongs to the concept question , Investigate the storage and representation of Chinese characters in computer . The input code refers to the code used for input that represents and encodes each Chinese character with a key ; Internal code is the binary code used to represent each Chinese character in the computer ; The national standard code refers to the standard code for each Chinese character ; The function of location code is similar to that of national standard code , However, there is a fixed difference between the national standard code and the information transmission . In a broad sense , Location code and national standard code belong to internal code , The inner code here is obviously the best option .

**7. Suppose the computer addresses in bytes , Small end mode is adopted , There is one float Type variable x The address for FFFF C000H start ,x=12345678H, be FFFF C001H is （C）**

A.1234H

B.34H

C.56H

D.5678H

analysis ： This question belongs to comprehension , Examine the storage of data in a computer . This question examines the fact that the computer addresses according to bytes , Therefore, each address should correspond to eight binary significant bits , That is, two hexadecimal digits , Thus eliminate AD. Because the computer adopts small terminal addressing , That is, the most significant bit is placed in the low position , So the starting address FFFF C000H Where is the 78H,FFFF C001H The contents stored in are 56H.

**8. The following character codes contain parity bits , Assuming no error has occurred , When odd check is adopted, the character code is （C）**

A.01010011

B.01100110

C.10110000

D.00110101

analysis ： This question belongs to comprehension , Examine the judgment method of parity test . The test bit result of odd test is the sum of each numerical bit and 1 The result of XOR operation , The test result is placed after the numerical digit （ In this question, it is the eighth bit of binary number ）. The simple calculation method of multiple number XORs is ： An odd number 1 The result is 1, Even number 1 The result is zero . Thus we can see that A The numeric bits of the options share 3 individual 1, And again 1 Do an XOR , The inspection result shall be 0, error ;B The option value bits are 4 individual 1, The inspection result shall be 1, error ;C The option value bits are 3 individual 1, The inspection result shall be 0, correct ;D The option value bits are 3 individual 1, The inspection result shall be 0, error .

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