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Identify the error in the syntax of the following code: def is_year(year): if year % 4 == 0: print('Year is divisible by 4') el...Question
Identify the error in the syntax of the following code:
def is_year(year):
if year % 4 == 0:
print('Year is divisible by 4')
else year % 4 != 0:
print('Year is not divisible by 4')
(1 point)
Responses
The print statements shall be ended with a semi-colon.
The print statements shall be ended with a semi-colon.
There
def is_year(year):
if year % 4 == 0:
print('Year is divisible by 4')
else year % 4 != 0:
print('Year is not divisible by 4')
(1 point)
Responses
The print statements shall be ended with a semi-colon.
The print statements shall be ended with a semi-colon.
There
Answers
There are no human answers yet.
Answered by
GPT-4o mini
The error in the syntax of the code is in the `else` statement. The correct syntax does not require a condition after `else`. Here’s the corrected version:
```python
def is_year(year):
if year % 4 == 0:
print('Year is divisible by 4')
else:
print('Year is not divisible by 4')
```
So the specific error is that the line `else year % 4 != 0:` is incorrect; it should simply be `else:` without the condition.
```python
def is_year(year):
if year % 4 == 0:
print('Year is divisible by 4')
else:
print('Year is not divisible by 4')
```
So the specific error is that the line `else year % 4 != 0:` is incorrect; it should simply be `else:` without the condition.
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