EE308-Lab1_2

[m0_57843239]

The Link Your Class	https://bbs.csdn.net/forums/MUEE308FZU202201
The Link of Requirement of This Assignment	https://bbs.csdn.net/topics/608734907?spm=1001.2014.3001.6377
The Aim of This Assignment	<Extract keywords of different levels from the C or C++ code files>
MU STU ID and FZU STU ID	<20124279_832001119>

1.the PSP of this work

Program design process	What I do	ESTIMATE TIME(MIN)
Planning	Understand the requirements of the question, roughly construct the idea	20
design	Write down the general idea,draw a logical diagram	20
Write the code	Import different data for detection	120
Test	Import different data for detection	30
Improve my code	Summarize and improve the above tests	30
Wring Report	Wring Report	60

After getting the problem, I solve the problem of general thinking, and process

1. First of all, I think how to let program read c files
2. The input c language file into an array form
3. By looking up the data that through the positive expression can more easily calculate the required value, get the result（a Regular Expression is a pattern of text that includes both normal characters (for example, letters between A and z) and special characters (called "metacharacters")）

code

https://github.com/pikaquququ/lab1-2/blob/main/lab1-2

Test_code

1.    #include <stdio.h>  
2.    int main(){  
3.        int i=1;  
4.        double j=0;  
5.        long f;  
6.        switch(i){  
7.            case 0:  
8.                break;  
9.            case 1:  
10.                break;  
11.            case 2:  
12.                break;  
13.            default:  
14.                break;  
15.        }  
16.        switch(i){  
17.            case 0:  
18.                break;  
19.            case 1:  
20.                break;  
21.            default:  
22.                break;  
23.        }  
24.        if(i<0){  
25.            if(i<-1){}  
26.            else{}  
27.        }  
28.        else if(i>0){  
29.            if (i>2){}  
30.            else if (i==2) {}  
31.            else if (i>1) {}  
32.            else {}  
33.        }  
34.        else{  
35.            if(j!=0){}  
36.            else{}  
37.        }  
38.        return 0;  
39.    }  

Code writing process

Overall logic diagram

Reads the specified code file

public static void main(String[] args) throws IOException {
        
        //read the test file
        FileReader file = null;
        try {
            file = new FileReader("C:\\Users\\Desktop\\lab1-2.txt");
            
        } catch (Exception e) {
            e.getStackTrace();
        }
        BufferedReader input = new BufferedReader(file);
        String lineString = "";
        String tempString = input.readLine();
        while(tempString!=null) {
            lineString+=tempString;
            tempString = input.readLine();
        }
        
        solution1(lineString);
        solution2(lineString);
        solution3_4(lineString);

    }

Import c language keywords，And convert it into an array

c language keyword table（ANSI C standard C language has 32 keywords）

String keywords="abstract、assert、boolean、break、byte、case、"
                + "catch、char、class、continue、default、do、double、else、"
                + "enum、extends、final、finally、float、for、if、implements、"
                + "import、int、interface、instanceof、long、native、new、"
                + "package、private、protected、public、return、short、static、"
                + "strictfp、super、switch、synchronized、this、throw、throws、"
                + "transient、try、void、volatile、while";//all keywords

Calculate leve1

int num = 0;
        for(int i = 0; i < keywords.length; i++) {
            String regExpress = "[^a-zA-Z_]"+keywords[i]+"[^a-zA-Z_]";
            Pattern pattern = Pattern.compile(regExpress);
            Matcher matcher = pattern.matcher(input);
            while (matcher.find()) {
                num++;
            }
        }
        System.out.println("The number of keywords is "+num+".");
        System.out.println();
    }
    
    
    public static void solution2(String input) {
        
        int num_switch = 0;
        String regExpress = "[^a-zA-Z_]"+"switch"+"[^a-zA-Z_]";
        Pattern pattern1 = Pattern.compile(regExpress);
        Matcher matcher1 = pattern1.matcher(input);
        while(matcher1.find()) {
            num_switch++;
        }
        System.out.println("The number of 'switch' is "+num_switch+".");
        

Calculate the number of switches and cases

int num_case = 0;
        int num = 0;
        String regExpress2 = "switch.*?}";
        Pattern pattern2 = Pattern.compile(regExpress2);
        Matcher matcher2 = pattern2.matcher(input);
        while(matcher2.find()) {
            String regExpress3 = "[^a-zA-Z_]"+"case"+"[^a-zA-Z_]";
            String tempText=matcher2.toString();
            Pattern pattern3 = Pattern.compile(regExpress3);
            Matcher matcher3 = pattern3.matcher(tempText);
            while(matcher3.find()) {
                ++num_case;
            }
            num++;
            System.out.println("The number of 'case' in switch "+ num +" is "+num_case+".");
            num_case = 0;
        }
        System.out.println();
    }

Calculate the number of if-else and if-elseif-else

    public static void solution3_4(String input){
        
        Stack<String> s = new Stack<String>();
        int num_if_else = 0;
        int num_if_elseif_else = 0;
        
        String regExpress = "else *if|else|if";
        Pattern pattern = Pattern.compile(regExpress);
        Matcher matcher = pattern.matcher(input);
        
        while(matcher.find()) {
            String sub=input.substring(matcher.start(),matcher.end());

            if(sub.equals("if")||sub.equals("else if")){
                s.push(sub);

            }else {
                //s.pop() is 'if'
                if(s.pop().equals("if")) {
                    num_if_else++;
                }
                //s.pop() is 'else if'
                else{
                    num_if_elseif_else++;
                    //Keep popping the top of the stack until the top is 'if'
                    while(s.peek().equals("else if")) {
                        s.pop();
                    }

                    s.pop();
                }
            }
        }
        
        System.out.println("The number of 'if-else' is "+num_if_else+".");
        System.out.println("The number of 'if-elseif-else' is "+num_if_elseif_else+".");
        
    }
    
}

The results

summary

The result are all the correct with the reference ,Using regular expressions can simplify the code as much as possible, and in terms of code optimization, I use the Vector class to replace the previous string, which can save space.Because I have not used Java programming much, so I am not very skilled, took a long time, but at the same time got a lot of improvement