### Break Down the Problem
1. We need to analyze the code provided and determine what happens when it is executed.
2. The function `xyz` is defined, and within it, a local variable `a` is assigned the value `56`.
3. After calling the function `xyz`, there is an attempt to print the value of `a`.

### Relevant Concepts
1. Understand the scope of variables in Python: variables defined inside a function are local to that function and cannot be accessed outside of it.
2. Recognize that an attempt to access a variable that is not in the global scope will raise a `NameError`.

### Analysis and Detail
1. The function `xyz()` sets a local variable `a = 56`.
2. After calling `xyz()`, the variable `a` defined inside `xyz` does not exist in the global scope.
3. The line `print(a)` is outside of the function and thus attempts to access the global variable `a`, which has not been defined.

### Verify and Summarize
Based on the understanding of variable scope in Python, the attempt to print `a` will result in a `NameError` since `a` is not available in the global scope.

### Final Answer
**NameError**

Question

### Break Down the Problem
1. We need to analyze the code provided and determine what happens when it is executed.
2. The function `xyz` is defined, and within it, a local variable `a` is assigned the value `56`.
3. After calling the function `xyz`, there is an attempt to print the value of `a`.

### Relevant Concepts
1. Understand the scope of variables in Python: variables defined inside a function are local to that function and cannot be accessed outside of it.
2. Recognize that an attempt to access a variable that is not in the global scope will raise a `NameError`.

### Analysis and Detail
1. The function `xyz()` sets a local variable `a = 56`.
2. After calling `xyz()`, the variable `a` defined inside `xyz` does not exist in the global scope.
3. The line `print(a)` is outside of the function and thus attempts to access the global variable `a`, which has not been defined.

### Verify and Summarize
Based on the understanding of variable scope in Python, the attempt to print `a` will result in a `NameError` since `a` is not available in the global scope.

### Final Answer
**NameError**

Knowee AI · Accepted Answer

### Break Down the Problem
1. We need to analyze the code provided and determine what happens when it is executed.
2. The function `xyz` is defined, and within it, a local variable `a` is assigned the value `56`.
3. After calling the function `xyz`, there is an attempt to print the value of `a`.

### Relevant Concepts
1. Understand the scope of variables in Python: variables defined inside a function are local to that function and cannot be accessed outside of it.
2. Recognize that an attempt to access a variable that is not in the global scope will raise a `NameError`.

### Analysis and Detail
1. The function `xyz()` sets a local variable `a = 56`.
2. After calling `xyz()`, the variable `a` defined inside `xyz` does not exist in the global scope.
3. The line `print(a)` is outside of the function and thus attempts to access the global variable `a`, which has not been defined.

### Verify and Summarize
Based on the understanding of variable scope in Python, the attempt to print `a` will result in a `NameError` since `a` is not available in the global scope.

### Final Answer
**NameError**

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

Question

Choose the Correct Answer(s)

Solution

Break Down the Problem

Relevant Concepts

Analysis and Detail

Verify and Summarize

Final Answer

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