I want to apply for Fiberhome Technology, and I want to know what I will take in the general written interview. Thank you ~ ~ hardware development.

I am a DSP engineer in written test. There is only one piece of paper and two pages in the paper. A total of 10 questions. The answer is on the white paper given.

1. What are fixed-point DSP chips and floating-point DSP chips? What are the advantages and disadvantages?

Fixed-point DSP and floating-point DSP are mainly distinguished according to the data format when DSP works. A DSP whose data works in fixed-point format is a fixed-point DSP, and a DSP whose data works in floating-point format is a floating-point DSP. Floating-point DSP can perform floating-point operation at one time, which is realized by hardware. Fixed-point DSP can not directly perform floating-point operations, and the completion of floating-point operations needs the help of programs.

Advantages: floating-point DSP has a larger operation range than fixed-point DSP. The addressing range of floating-point DSP is also larger. The precision of floating-point DSP is higher. Floating-point operation is very fast.

Disadvantages: Fixed-point DSP needs to pay attention to the overflow of operation at all times. The accuracy is not enough, and the fixed-point DSP operation speed is slow.

2.2 What is the biggest difference? DSP and general CPU? What do they have in common?

(1) First of all, the architecture: the CPU is Von Neumann architecture, while the DSP has independent code and data bus, namely "Harvard architecture".

(2) Standardization and generalization: the standardization and generalization of CPU are well done and support the operating system, so the system with CPU as the core is convenient for human-computer interaction and communication with standard interface devices, which is very convenient and does not require hardware development; But it also makes the interface circuit of CPU peripherals more complicated. DSP is mainly used to develop embedded signal processing system, which does not emphasize human-computer interaction and generally does not need many communication interfaces, so its structure is relatively simple and easy to develop. If we only focus on embedded applications, the difference between embedded CPU and DSP should be that one focuses on control and the other on operation.

(3) Pipeline structure: Most DSPs have pipeline structure, that is, each instruction is fetched, decoded, fetched and executed by multiple functional units on the chip, which can greatly improve the execution efficiency of the system. However, the adoption of pipeline also increases the difficulty of software design, requiring designers to consider the requirements of pipeline in program design.

Fast multiplier: Multiply-accumulate (MAC) operation is often used in signal processing algorithms. DSP has a dedicated hardware multiplier, which can complete the MAC operation in one clock cycle. Hardware multiplier occupies a large area of DSP chip. (On the contrary, the general CPU uses slow iterative multiplication technology, which can complete a multiplication operation in multiple clock cycles, but it takes up less silicon resources).

Address generator: DSP has a dedicated hardware address generation unit, so it can support specific data address patterns required by many signal processing algorithms. This includes pre (post) addition (subtraction), modular address of ring data buffer and bit inversion address of FFT. The address generator unit works in parallel with the main ALU and multiplier, which further increases the workload that DSP can complete in one clock cycle.

Hardware-assisted loop: signal processing algorithms usually need to execute tight instruction loops. The support of hardware-assisted loop allows DSP to effectively execute code blocks without stopping the pipeline or letting software test loop termination conditions.

3. What's the difference between preemptive system and non-preemptive system? Which is better, DSP/BIOS or ucOS?

Pre-emptive, if the task with higher priority enters the ready state during the running of the current task, the current task will immediately hand over the CPU to the task with higher priority.

Nonpreemptive: Wait until the current task is completed before handing it over to the task with higher priority.

DSP/BIOS and ucOSII are preemptive operating systems.

4. Talk about your understanding of DSP map.

Mapping file is a mapping file of all programs, data and IO spaces used by DSP after CCS software is compiled.

First, the generation method

There are two main methods to generate map files. One is automatically generated by the system, and the default file name is the established project name (such as XXX is the project name) plus XXX. Map suffix. The other is to specify the generation of the mapping file in the CMD file. The operation method is to add "-mabc". Map "is placed before the memory instruction, and the file name can be arbitrary.

Second, the file format

The mapping file is roughly divided into four parts: file header, memory configuration, segment mapping and global symbol. The memory configuration is related to the memory instructions in the CMD file, and the program and data interval definitions defined in the CMD file can be found in this section. When it is different from the CMD file, it will add an actually used interval in the mapping file, that is, the length of space actually used in the program. The segment mapping part is associated with the SECTION instruction in the CMD file. In this part of the program, the starting address and actual length of all segment mappings are specified in detail. It can be specific to the segment specified by PROGMA in the program and the OBJ file generated by each individual file. The global symbol can be every function used by the program, and the global variable defined in the program can find the corresponding starting address here. Map files are arranged in two ways, one is by corresponding address, and the other is by global symbol alphabet.

5. Simple procedural issues. Look at the program, write the output, written in C.

6. What is the two's complement of 10 for-7 and10? Write Q 15 0.25 and Q 15 0.5.

(1) The complement of a positive number is the same as the original code.

(2) The complement of a negative number is the inversion of its original code bit by bit, except the sign bit; Then add 1 to the integer.

The original code of -7 converted into binary is100000011,and the inverse code is11165438+. The complement is the complement+1=11111165438+.

The original binary code of 10 is 0000000010/0, which is a positive number, and its complement is the same as the original code.

The number of DSP chips is expressed in the form of two's complement. Each number of 16 is represented by a sign bit, with 0 representing a positive value and l representing a negative value. The remaining 15 bits indicate the size of the value.

For example, the binary number 00100000001b = 8195.

Binary number11111111.

There are two kinds of digital calibration: Q representation and S representation.

The numbers represented by different Q's not only have different ranges, but also have different precision. The larger the q, the smaller the numerical range, but the higher the accuracy;

For example, the binary number 16 is 2000H=8 192, which is represented by Q0.

16 binary number 2000H=0.25, expressed by Q 15.

7. Talk about the main steps of DSP development.

8. Write the function prototype of strcpy.

char* strcpy(char *strDest，const char * strSrc)

{

}

9. Write a function to create a single linked list.

Structure stu_info{

int x；

struct stu _ info * next

};

void inputs tuinfo(struct stu _ info * * head，int n)

{

}

10. Rearrange a string so that letters come first, numbers come last, and the relative positions of letters and numbers remain unchanged.

Including & ltstdio.h & gt

Include & ltstring.h & gt

Including & ltstdlib.h & gt

Function name: fun

Parameters: char*s, int *m

Function return value: chLetter (array name of array chLetter[N])

Local variable: charchletter [n]; char ch number[N]； int i，j，k；

char *fun(char*s，int *m)

{