Greetings, I recently started coding and there's an error I'm getting which I have posted below the code. Your help in resolving the error will be highly appreciated.

Question

1.00/5 (1 vote)

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Python

list=[]
print(''' press 1 for entering no. into stack
          press 2 for deleting no. from stack  
          press 3 for viewing last no. in stack
          press 4 for viewing entire stack
''')
a=int(input("enter your choice: "))
if a==1:
       b=int(input("enter no. of elements you wanna insert in stack: "))
       for i in range(b):
             c=int(input("enter element " + str(i+1) + ": "))
             list.append(c)
print(list)
print()
elif a==2:
       e=int(input("enter no. of elements you wanna delete from stack: "))
       if len(list)==0:
           print("empty stack: ")
       elif e < len(list):
           print("insufficient elements in list to delete: ")
       else:
           for i in range(e):
               f=list.pop()
               print("popped element is: ", f)
               print("list after popping element :", list)
           print()
           print("final list after popping:", list)
print()
elif a==3:
        if len(list) == 0:
           print("empty stack: ")
           print()
        else:
           print("last element in stack is : ", list[-1])
           print()
elif a==4:
        if len(list) == 0:
           print("empty stack: ")
           print()
        else:
            print("list is: ", list)
            print()
elif a==5:
        print("u have exited......")
        print()
        break;
else:
        print("invalid input")

__________________________________________


ERROR:

 elif a==2:
       ^
SyntaxError: invalid syntax

What I have tried:

I looked it up on google and checked the syntax but couldnt resolve the issue. Your help will be highly beneficial for me and would be much appreciated.

Posted 26-Jul-22 20:42pm

akshaychauhan0101

Updated 26-Jul-22 21:36pm

CPallini

v2

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3 solutions

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CPallini · Accepted Answer · 2022-07-26T21:01:00

the elif block must immediately follow an if (or another elif) one. For example, this is allowed

Python

if i<5:
  print("i is less than 5")
elif i<20:
  print("i is between 5 and 19")
else:
  print("i greater thn 19")

This is allowed as well

Python

if i<5:
  print("i is less than 5")
  print(i)
elif i<20:
  print("i is between 5 and 19")
else:
  print("i greater thn 19")

This is NOT allowed

Python

if i<5:
  print("i is less than 5")
print(i) # the if block has completed before this statement
elif i<20: # this elif doesn't follow an if block
  print("i is between 5 and 19")
else:
  print("i greater thn 19")

because, when execution reached the print(i) statement (note its indentation), the if block has already completed.

OriginalGriff · Accepted Answer · 2022-07-26T21:21:00

To add to what CPallini has - rightly - said, in Python indentation is very important - it controls what is and isn't in a block of code.
A simple block of code must all have the indentation the same: the first line that has less spaces before it ends the block.

So in your code:

Python

if a==1:
       b=int(input("enter no. of elements you wanna insert in stack: "))
       for i in range(b):
             c=int(input("enter element " + str(i+1) + ": "))
             list.append(c)
print(list)
print()
elif a==2:

Teh two print statements end the if block, and the following elif has no if to match up with.
Indent it, and it works again:

Python

if a==1:
       b=int(input("enter no. of elements you wanna insert in stack: "))
       for i in range(b):
             c=int(input("enter element " + str(i+1) + ": "))
             list.append(c)
       print(list)
       print()
elif a==2:

Because the print lines are part of the conditional code block.

Patrice T · Accepted Answer · 2022-07-26T21:36:00

Quote:
there's an error I'm getting which I have posted below the code.

In Python, indentation is the programm structure.

Python

if a==1: # this if
       b=int(input("enter no. of elements you wanna insert in stack: "))
       for i in range(b):
             c=int(input("enter element " + str(i+1) + ": "))
             list.append(c)
print(list) # ends here
print()
elif a==2: # thus, this is not linked to previous if

But this is ok

Python

if a==1:
       b=int(input("enter no. of elements you wanna insert in stack: "))
       for i in range(b):
             c=int(input("enter element " + str(i+1) + ": "))
             list.append(c)
       print(list)
       print()
elif a==2:

if and elif are at same level and there is nothing else at same level in between.

Greetings, I recently started coding and there's an error I'm getting which I have posted below the code. Your help in resolving the error will be highly appreciated.

3 solutions

Solution 1

Solution 2

Solution 3

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