2.14 LAB: Exception handling to detect input string vs. integer   The given program reads a list of single-word first names and ages (ending with -1), and outputs that list with the age incremented. The program fails and throws an exception if the second input on a line is a string rather than an integer. At FIXME in the code, add try and except blocks to catch the ValueError exception and output 0 for the age. Ex: If the input is: Lee 18 Lua 21 Mary Beth 19 Stu 33 -1 then the output is: Lee 19 Lua 22 Mary 0 Stu 34

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
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2.14 LAB: Exception handling to detect input string vs. integer

 

The given program reads a list of single-word first names and ages (ending with -1), and outputs that list with the age incremented. The program fails and throws an exception if the second input on a line is a string rather than an integer. At FIXME in the code, add try and except blocks to catch the ValueError exception and output 0 for the age.

Ex: If the input is:

Lee 18 Lua 21 Mary Beth 19 Stu 33 -1

then the output is:

Lee 19 Lua 22 Mary 0 Stu 34
G entry level
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Content
zy Section 2.14 X
(229) ITS12 X
G 2.14 LAB: E X
b Answered:
* (Solved] In
Contact Us
M Inbox (7,71 ×
+
i learn.zybooks.com/zybook/CYB_135_54915392/chapter/2/section/14
= zyBooks My library > CYB/135: Object-Oriented Security Scripting home > 2.14: LAB: Exception handling to detect input string vs. integer
E zyBooks catalog
? Help/FAQ
Kenneth Schultz -
then the output is:
Lee 19
Lua 22
Mary 0
Stu 34
346682.2019644.gx3zgy7
LAB
2.14.1: LAB: Exception handling to detect input string vs. integer
0/ 10
АCTIVITY
main.py
Load default template..
1 # Split input into 2 parts: name and age
2 parts = input().split()|
3 name = parts[0]
4 while name != '-1':
# FIXME: The following line will throw ValueError exception.
Insert try/except blocks to catch the exception.
age = int(parts[1]) + 1
print('{} {}'.format(name, age))
7
8
9
10
# Get next line
parts = input().split()
name = parts[0]
11
12
Run your program as often as you'd like, before submitting for grading. Below, type any needed
input values in the first box, then click Run program and observe the program's output in the
second box.
Develop mode
Submit mode
Enter program input (optional)
Lee 18
6:05 AM
O Type here to search
75°F Rain to stop
9/16/2021
Transcribed Image Text:G entry level * Home Content zy Section 2.14 X (229) ITS12 X G 2.14 LAB: E X b Answered: * (Solved] In Contact Us M Inbox (7,71 × + i learn.zybooks.com/zybook/CYB_135_54915392/chapter/2/section/14 = zyBooks My library > CYB/135: Object-Oriented Security Scripting home > 2.14: LAB: Exception handling to detect input string vs. integer E zyBooks catalog ? Help/FAQ Kenneth Schultz - then the output is: Lee 19 Lua 22 Mary 0 Stu 34 346682.2019644.gx3zgy7 LAB 2.14.1: LAB: Exception handling to detect input string vs. integer 0/ 10 АCTIVITY main.py Load default template.. 1 # Split input into 2 parts: name and age 2 parts = input().split()| 3 name = parts[0] 4 while name != '-1': # FIXME: The following line will throw ValueError exception. Insert try/except blocks to catch the exception. age = int(parts[1]) + 1 print('{} {}'.format(name, age)) 7 8 9 10 # Get next line parts = input().split() name = parts[0] 11 12 Run your program as often as you'd like, before submitting for grading. Below, type any needed input values in the first box, then click Run program and observe the program's output in the second box. Develop mode Submit mode Enter program input (optional) Lee 18 6:05 AM O Type here to search 75°F Rain to stop 9/16/2021
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