The output will be a NameError.

Here's why:

1. The function `xyz()` is defined, and within this function, a local variable `a` is set to 56.

2. The function `xyz()` is called. This executes the function and sets the local variable `a` to 56. However, this variable `a` only exists within the scope of the function `xyz()`.

3. After the function call, `print(a)` is executed. However, `a` is not defined in this scope (outside of the function), so Python will raise a NameError, indicating that `a` is not defined.

So, the correct option is NameError.

Question

The output will be a NameError.

Here's why:

1. The function `xyz()` is defined, and within this function, a local variable `a` is set to 56.

2. The function `xyz()` is called. This executes the function and sets the local variable `a` to 56. However, this variable `a` only exists within the scope of the function `xyz()`.

3. After the function call, `print(a)` is executed. However, `a` is not defined in this scope (outside of the function), so Python will raise a NameError, indicating that `a` is not defined.

So, the correct option is NameError.

Knowee AI · Accepted Answer

The output will be a NameError.

Here's why:

1. The function `xyz()` is defined, and within this function, a local variable `a` is set to 56.

2. The function `xyz()` is called. This executes the function and sets the local variable `a` to 56. However, this variable `a` only exists within the scope of the function `xyz()`.

3. After the function call, `print(a)` is executed. However, `a` is not defined in this scope (outside of the function), so Python will raise a NameError, indicating that `a` is not defined.

So, the correct option is NameError.

What will be the output after the following statements?def xyz(): a = 56 xyz() print(a)OptionsNameErrora = 56xyz56

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