The given code attempts to perform a division operation that will raise an exception, but it is trying to catch a `ValueError`, which is not the correct exception for a division by zero in Python. Here's a breakdown of how the code will execute:

1. **Line Analysis**:
- `a = 10`: This sets the variable `a` to 10.
- `b = 0`: This sets the variable `b` to 0.
- `c = a/b`: Division by zero occurs here, which raises a `ZeroDivisionError`.
- The `print(c)` statement is never reached because of the exception raised.

2. **Exception Handling**:
- The `except ValueError` block is meant to catch a `ValueError`, but because a `ZeroDivisionError` is raised instead, this block will not execute.

3. **Conclusion**:
- Since the specified exception type does not match the raised exception, the program will terminate with a runtime error and print a traceback indicating a `ZeroDivisionError`.

### Final Answer
The output of the code will be a runtime error, specifically `ZeroDivisionError: division by zero`. The message "Exception occurred" will not be printed.

Question

The given code attempts to perform a division operation that will raise an exception, but it is trying to catch a `ValueError`, which is not the correct exception for a division by zero in Python. Here's a breakdown of how the code will execute:

1. **Line Analysis**:
   - `a = 10`: This sets the variable `a` to 10.
   - `b = 0`: This sets the variable `b` to 0.
   - `c = a/b`: Division by zero occurs here, which raises a `ZeroDivisionError`.
   - The `print(c)` statement is never reached because of the exception raised.

2. **Exception Handling**:
   - The `except ValueError` block is meant to catch a `ValueError`, but because a `ZeroDivisionError` is raised instead, this block will not execute.

3. **Conclusion**:
   - Since the specified exception type does not match the raised exception, the program will terminate with a runtime error and print a traceback indicating a `ZeroDivisionError`.

### Final Answer
The output of the code will be a runtime error, specifically `ZeroDivisionError: division by zero`. The message "Exception occurred" will not be printed.

Knowee AI · Accepted Answer

The given code attempts to perform a division operation that will raise an exception, but it is trying to catch a `ValueError`, which is not the correct exception for a division by zero in Python. Here's a breakdown of how the code will execute:

1. **Line Analysis**:
   - `a = 10`: This sets the variable `a` to 10.
   - `b = 0`: This sets the variable `b` to 0.
   - `c = a/b`: Division by zero occurs here, which raises a `ZeroDivisionError`.
   - The `print(c)` statement is never reached because of the exception raised.

2. **Exception Handling**:
   - The `except ValueError` block is meant to catch a `ValueError`, but because a `ZeroDivisionError` is raised instead, this block will not execute.

3. **Conclusion**:
   - Since the specified exception type does not match the raised exception, the program will terminate with a runtime error and print a traceback indicating a `ZeroDivisionError`.

### Final Answer
The output of the code will be a runtime error, specifically `ZeroDivisionError: division by zero`. The message "Exception occurred" will not be printed.

What will be the output of the following code?try: a = 10 b = 0 c = a/b print(c)except ValueError: print(“Exception occurred”)

Question

What will be the output of the following code?

Solution

Final Answer

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