EE308 LAB1-2

1. PSP TABLE

2. PROBLEM-SOLVING IDEAS

• Subject Analysis
  The questions have four levels of requirements, each of which increases in difficulty, and the latter level includes the requirements of the previous level. So the inspiration for me is that when coding, you can start at the front and gradually increase the code to achieve higher level requirements.

• Keyword Search
  To retrieve the keyword content of a file, we read the file into a string or vector and use the string::find() function to retrieve the file. The key and most important issue is to make sure that the keyword retrieved is a keyword, that is, you cannot select the same part of the variable name, comment, function name as the keyword.

• Choice of Language —— Python
  The first question is the choice of language. I briefly review the languages I have learned. I first learned Python, and then I learned C and C++. Among several languages, the one I am most familiar with is Python, because I have recently learned it and am proficient in using it. Another reason is that there are relatively many libraries to call in Python, and the code is relatively simple to implement.

• Fixed Structure Search
  This is followed by quantitative lookups for switch-case, if-else, and if-elseif-else constructs.I learned pushing、 popping and so on, through the establishment of stack count.

3. THE FLOW CHART

4. CODE DESCRIPTION

Ⅰ. Read File

f = open(r"D:\daima.txt","r",encoding="utf-8") # open file
data = f.read() # read file

Ⅱ. Count the Number of Keywords

data = re.sub(r"\"[\S\s]*?\"", "", data)  # Delete comments and quotes
data = re.sub(r"\/\*[\S\s]*?\*\/", "", data)
res = re.findall(r'\bdo\b',data) # Look up the list of do's from data
total += len(res) # count the number of DO's by LEN

Ⅲ. Count the Number of Switches

This part is same to the part 2

Ⅳ. Count the Number of Cases in each set of Switch Structures

Check Switch first and start checking cases and counting. If you encounter default or the next Switch or the file is finished, stop and save the number of cases.

i = 0
k = len(f.readlines())  # Total number of rows
f.seek(0)
for i in range(k):
    data = f.readline().strip()  # Read line by line 
    data = re.sub(r"\"[\S\s]*?\"", "", data)
    data = re.sub(r"\/\*[\S\s]*?\*\/", "", data)
 
    res = re.findall(r'\bswitch\b', data)
    if flag == 1 and len(res) > 0:  # The next switch structure is encountered
        flag = 2
    if len(res) > 0 and flag == 0:  # Find the switch structure
        flag = 1
    if flag == 1:  
        res = re.findall(r'\bcase\b', data)
        t2 = t2 + len(res)
        res = re.findall(r'\bdefault\b', data)
        if len(res) > 0:  # exit at default
            flag = 0
            arva[x] = t2  # Save the number of case
            x = x + 1
            t2 = 0  # t2 reset
    if flag == 2:  # meet the next switch structure
        flag = 1  # Prepare to count the number of cases for the next switch structure
        arva[x] = t2  # Save the number of case
        x = x + 1
        t2 = 0  # t2 reset
    if i == k - 1 and t2 != 0:  # There is no default and switch
        arva[x] = t2  # Save the number of case
        x = x + 1
        t2 = 0  # t2 reset

Ⅴ. Count the number of if-else structures and if-elseif-else structures

Met with elseif if 1 stack, stack 2, meet {stack 3, meet the else judge which structure belongs to the else, met} see whether ordinary braces can delete directly, or used to control the level of the level of finish can match the match can not match the delete to delete the curly braces.

 
    res = re.findall(r'\bif\b', data)
    res1 = re.findall(r'\belse\b', data)
    res2 = re.findall(r'{', data)
    res3 = re.findall(r'}', data)

    if (len(res) > 0) and (len(res1) == 0):  # Only if with NO else
        st.push(1)  # 1 push
    if (len(res) > 0) and (len(res1) > 0):  # if AND else
        st.push(2)  # 2 push
    if (len(res1) > 0) and (len(res) == 0):  #ONLY else NO if
        if st.isEmpty() == False and st.peek() == 2:  
            t4 = t4 + 1
            while st.peek() == 2:  # delete all of the elseif
                st.pop()
            st.pop()  # delete if
        if st.isEmpty() == False and st.peek() == 1:  
            t3 = t3 + 1
            st.pop()  # delete if
    if (len(res2) > 0):
        st.push(3)  
    if (len(res3) > 0):  
        if st.isEmpty() == False and st.peek() == 3:
            st.pop()  
        elif (st.isEmpty() == False and st.peek() == 2) or (st.isEmpty() == False and st.peek() == 1):
            while st.peek() != 3: 
                st.pop()
            st.pop()  

5. PRESENT THE OUTCOME

6. SUMMARY

Compared with the simple coding tasks assigned by the teacher before, this experiment was a great challenge, because it was also the first time for me to try to complete a difficult programming task from beginning to end. I encountered many difficulties in the process, but I successfully solved them through continuous learning of relevant knowledge. I hope that I can keep this spirit of not afraid of difficulties in my future study, and keep learning and improving myself.