Question

Python3 command line arguments; there are unknown number of arguments, 'only number // number and the length of the argument is three' is a valid argument,for example,2//5,3//5 are valid arguments,a//2,b//4,7//0,b//k are not valid arguments, calculate each valid argument and the number of valid arguments, and then print

hint: split is very useful

restrictions: [1:], for loop, slicing, lambda function, 'in' keyword,isdigit(), isalpha() cannot be used. Exception handling must be used, it is recommended to use while loop to solve this problem.

Cuser@sahara ]$ python Q3.py 2//0 a//b 3//a b//2 8//3 9//4 5/2 8//3 2 9//4 2 There are 2 valid arguments Cuser@sahara ~] python Q3.py 5//2 4//3 7//2 5//2 2 4//3 1 7//2 3 There are 3 valid arguments Cuser@sahara

Answer 1

Here is the answer...

CODE:

import sys
def find(s):
i=0
#print(s)
while(i<len(s)):
if s[i]=='/':
return i
i+=1
def join(l):
i=0
s=""
while(i<len(l)):
s+=l[i]+" "
i+=1
return s
length=len(sys.argv)
sys.argv=join(sys.argv)
a=sys.argv.split(" ")
#print(a)
index=1
l=[]
while(index<length):
l.append(a[index]) #append each argument to list
index+=1

j=0
count=0 #to take 3 number// number
valid=0 #for valid count
while(j<len(l)): #iterate through list of numbers
i=0
#print(l[j])
k=find(l[j]) #find the // operator in argument
#print(k)
if(count==3): #if count 3 break
break
s1="" #for storing first number
while(i<k):
s1+=l[j][i] #concatenate to string
i+=1
i+=2
#print(i)
s2="" #for denominator/second number
while(i<len(l[j])):
s2+=l[j][i] #concatenate to second string
i+=1
#print(s1,s2)
j+=1 #for next argument
try:
a=int(s1) #convert first number to integer
b=int(s2) #convert second number to integer
#print(a,b)
count+=1 #count the number//number arguments
#if possible then print and count valid arguments
print(a,"//",b," = ",a//b)
valid+=1
  
except:
print(end="") #cathing the raised exception and do nothing
print("There are ",valid," valid arguments") #print valid arguments
  
  
  

  

CODE Snapshot:

- 0 X 3 File Users\sai Desktop 03.py - Sublime Text (UNREGISTERED) Edit Selection Find View Goto Tools Project Preferences Help SWEEVOUW Q3.py import sys def find(s): i=0 #print(s) while(i<len(s)): if s[i]=='/': return i i+=1 def join(L): i 15 16 while(i<len(1)): s+-l[i]+" " i+-1 returns length=len(sys.argv) sys.argv=join(sys.argv) a sys.argv.split(" ") 19 #print(a) 20 index=1 21 1-[] 22 while(index< length): 23 1.append(a[index]) 24 index=1 #append each argument to list 26 28 j- count=0 valid=0 while(j<len(1)): i @ #print(1[j]) k=find(1[j]) #to take 3 number// number #for valid count #iterate through list of numbers 30 #find the // operator in argument

s1- $2 " k-find(1[j]) #find the // operator in argument #print(k) if(count==3): #if count 3 break break #for storing first number while(i<k): s1+=1[j][i] #concatenate to string i+=1 i+=2 #print(i) #for denominator/second number while(i<len(1[j])): s2+=1[j][i] #concatenate to second string i+=1 #print(s1, s2) j+=1 #for next argument try: a=int(s1) #convert first number to integer b=int(52) #convert second number to integer #print(a,b) count+=1 #count the number//number arguments #if possible then print and count valid arguments print(a,"//",b," = ",a//b) valid+=1 except: print(end="") #cathing the raised exception and do nothing 59 print("There are ",valid," valid arguments") #print valid arguments

OUTPUT:

C:\WINDOWS\system32\cmd.exe - O X C:\Users\sai\Desktop>python 03.py_27/0 a/b 3//a b//28/73 9//4 5/2 8 // 3 = 2 9 // 4 = 2 There are 2 valid arguments C:\Users\sai\Desktop>python Q3.py 5//24/73 77/2 5 // 2 = 2 4 // 3 = 1 7 // 2 = 3 There are 3 valid arguments C:\Users\sai\Desktop>python Q3.py 3/75 2774 3 // 5 = 0 2 // 4 = 0 There are 2 valid arguments C:\Users\sai\Desktop>

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