EE308_LAB2

Miracle_Han 2021-09-23 20:07:06

The Link Your Class	https://bbs.csdn.net/forums/MUEE308FZ?category=0
The Link of Requirement of This Assignment	https://bbs.csdn.net/topics/600798588
The Aim of This Assignment	Read keywords and statement structures in files and learn to use Git
MU STU ID and FZU STU ID	19104375 & 831902106

0. Prerequisites

0.1 Git

Before we learn to use Git, what do we need to know about Reversion control.

0.1.1 What is Revision Control ?

Definition: Revision control is a software engineering technique used during development to manage the history of changes made to files, directories, or projects. It can be easily viewed and backed up to restore previous versions.

It can help us:

Realize cross-region multi-person collaborative development
To track and record the history of one or more files
Organize and protect your source code and documentation
Statistical workload
Parallel development, improve development efficiency
Track and record the entire software development process
Reduce developer burden, save time, and reduce human error

0.2 basic operations

git add . # add the babababab

git commit -m"comment" # put the comment such as time, author in the comment block

git push origin main

download Git on https://git-scm.com/

Here's my Github screenshot
My Github Adress: https://github.com/Miracle-Han/EE308

1. PSP From of this LAB

Personal Software Process Stages	Estimated Time/minutes	Completed Time/minutes
Planning	30	45
Estimate	20	30
Development
Analysis	60	120
Design Spec	20	60
Design Review	5	10
Coding Standard	40	60
Design	30	120
Coding	240	600
Code ReviewPlanning	30	60
Test	30	60
Reporting	60	100
Test Report	60	40
Size Measurement	15	30
Postmortem&Process Improvement	60	30
total	670	1265

2. Problem-solving Ideas

2.1 Program Requirements

Basic requirement: output "keyword" statistics
Advanced requirement: output the number of "switch case" structures, and output the number of "case" corresponding to each group
Uplifting requirement: output the number of "if else" structures
Ultimate requirement: output the number of "if, else if, else" structures

Before completing the more difficult requirements, you need to complete the Lower requirements.

2.2 Personal Idea

I plan to use C++ to complete this experiment
Based on this LAB, I decided to divide it into three parts

IO
Split the text into keywords
Find and output

3. Implementation process

3.0 Preparatory work

Before we type "int main", we need to prepare the following things.

Head file

#include <iostream>
#include <string>
#include <fstream>
#include <sstream>
using namespace std;

Some of the parameters

/*for switch*/
struct CaseNum{ 
    bool activate_switch=false; 
    bool activate_brace=false; 
};

/*for case*/
CaseNum c[100];      
static int structNum =0; 
int caseArr[100] = {0};
static int casePos=0;

/*for if*/
struct The_if{
    bool hasIf = false;
    bool elseIf = false;
    bool hasElse= false;
};

The_if ifArr[100];
int ifArrPos = -1;
static int IE=0, IEE=0;

File Input

The entire file needs to be read first. Use c++ file input output stream, through the Internet access to the relevant information to master the relevant usage of ifstream.

Here is the code:

char file_add[100];
cin >> file_add;
ifstream cpp_file(file_add);

Processing comments and other information

The second is to read the contents of the file. Here getLine is used to read the contents of the file line by line and transpose unused strings to Spaces. However, before we start, we need to determine if there is a next row.

getline(fstream, str); // Make the content of the file into a string.

Here is the code:

while(!cpp_file.eof()){         
    getline(cpp_file, line); 
        for(int i=0;i<line.length();i++){
            if( ! (line[i]>='a' && line[i]<='z' || line[i]>='A'&&line[i]<='Z' || line[i]>='0' && line[i]<='9' || line[i] == '{'                  || line[i]=='}' ||  line[i]=='_' ||line[i]=='/' || line[i]=='"'))
              line[i]=' ';    
      } 
    /* For later convenience, we firstly transpose unused strings to Spaces. */
    
    string str;
    istringstream is(line); /* Read a separate string in each line*/
}

3.1 Requirements 1 - Keywords

3.1.1 Keywords list and Store code

auto	break	case	char	const	continue	default	do
double	else	enum	extern	float	for	goto	if
int	long	register	return	short	signed	sizeof	static
struct	switch	typedef	union	unsigned	void	volatile	while

const string Keywords_list[32] = {"auto", "break",    "case",    "char",    "const", "continue","default",    "do",
                            "double",    "else",    "enum",    "extern", "float", "for", "goto" , "if",
                            "int",    "long",    "register",    "return", "short", "signed", "sizeof", "static",
                            "struct",    "switch",    "typedef",    "union", "unsigned", "void", "volatile", "while"};

3.1.2 The number of Keywords

In this section, we need to consider the interference of invalid information, i.e. the non-keyword part contains keyword fragments
So I tested the length of each string, and only if the length and information were the same as the list keyword, Keywords_num++.

Here is the code:

while(is >> str){
        if(str[0]=='"'||str[0] == '/') continue; 
        /*Exclude comments and "double-quoted" text from the result*/
        for(int i=0; i < 32; i++){ 
            if( str.length() == keylen[i] ){
                if( str.compare(Keywords_list[i])==0 )
                    Keywords_num++;
        } 
        /*In this forLoop, if the length and information were the same as the list keyword, Keywords_num++, and we can get the total number of the keywords.*/
}

3.2 Requirements 2 - switch and case

Since the nesting of switches is not considered, the processing mechanism is simpler. The group number is determined by looking for the number of switches, and the number of cases before the next switch appears is counted as the number of cases in the group.

Here is the code:

while(is >> str){
    if(str[0]=='"'||str[0] == '/') continue;
    /*Exclude comments and "double-quoted" text from the result*/
    
    if( str.compare(Keywords_list[i])== 0 && i==25)
        c[structNum].activate_switch=true;
    
    if( str.compare("{")==0 && c[structNum].activate_switch==true)
        c[structNum].activate_brace=true;
    
     if(c[structNum].activate_switch && c[structNum].activate_brace ){
        if( str.compare(Keywords_list[i])==0 && i==2) /*Only when both are active can the count begin*/
            caseArr[casePos]++;
    }
 
    if(c[structNum].activate_switch && c[structNum].activate_brace && str.compare("}")==0){
        structNum++;
        casePos++;    
    }
}

3.3 Requirements 3 & 4 - if else and if else_if else

Get information for each "if" group using find and array of structs
Because "else if" has "else" and "if", so "else if" must be tested first.

Here is the code:

while(is >> str){
    if(str[0]=='"' || str[0]=='/')   continue;
    /*Exclude comments and "double-quoted" text from the result*/
    
    if( line.find("else if")>=0 && line.find("else if")<line.length()){
        ifArr[ifArrPos].elseIf = true;
    } else if( line.find("if")>=0 && line.find("if")<line.length()){
    ifArrPos++;
    
    while(ifArr[ifArrPos].hasElse && ifArrPos<100）
        ifArrPos++;
        
    ifArr[ifArrPos].hasIf = true;
       } else if ( line.find("else")>=0 && line.find("else")<line.length()) {
        ifArr[ifArrPos].hasElse = true;
        ifArrPos--;
        while(ifArr[ifArrPos].hasElse==true && ifArrPos>=0)
            ifArrPos--;
         }
}

Identify different if types, such as "if else" "if else_if else", The respective numbers of the two types of IF structures are iterated through the for loop

The_if ifArr[100];
int ifArrPos = -1;
static int IE=0, IEE=0;

for(int i=0;i<100;i++){
        if(ifArr[i].hasIf){
            if(ifArr[i].elseIf) IEE++;
            else{   IE++;
            }
        }
}

4. Program running result

Read the sample file and set the extraction level to a maximum of 4, resulting in the following output.

5. Unit test screenshots and description.

Using Visual Studio to measure the Proference

6. Summary

I haven't used c++ programming for a long time, and it was a bit intermittent at the beginning of writing, so this experiment was very challenging for me. This assignment is relatively simple if it judges the more standard code like the sample, but it is very difficult if it meets a special case and requires many special cases. Therefore, some special if else_if else structure judgment still exists some problems by the deadline of assignment. Through this assignment, I learned a lot of things that I had never used before, such as Git, Typora, Github... All in all, the experiment was very rewarding and rewarding.

And I want the next experiment to be easier!

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