Question at position 20 Which of the following lines of code will run, but then give a runtime error?dict = {"a": 3, "b": 2, "c":7}print(dict)dict = {"a": 3, "b": 2, "c":7}print(dict[a])dict = {"a": 3, "b": 2, "c":7}print(dict["a"])dict = {"a": 3, "b": 2, "ab":7}print(dict["a"+"b"])Clear my selection

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BlackTom AI · Accepted Answer

Using an undefined identifier as a dictionary key causes a NameError at runtime, even though the preceding code (dictionary creation) executes successfully.

1) Relevant concepts
- In Python, dictionary keys are looked up using the evaluated expression inside the brackets: dict[key_expr].
- A bare name like a is treated as a variable/identifier, not a string literal.
- If that variable is not defined, evaluating the expression raises NameError at runtime.
- Assigning to the name dict is allowed (it shadows the built-in name) and is not a runtime error by itself.

2) Step-by-step reasoning for each option
Option A: dict = {"a": 3, "b": 2, "c":7}
print(dict)
- Create the dict object successfully.
- print(dict) prints the dictionary representation. No error.

Option B: dict = {"a": 3, "b": 2, "c":7}
print(dict[a])
- Create the dict object successfully.
- When evaluating print(dict[a]), Python evaluates the index expression a. Since a is an undefined variable, evaluation fails and raises NameError. The program runs up to this point and then errors at runtime.

Option C: dict = {"a": 3, "b": 2, "c":7}
print(dict["a"])
- The key expression is the string literal "a". The lookup succeeds and prints 3. No error.

Option D: dict = {"a": 3, "b": 2, "ab":7}
print(dict["a"+"b"])
- The expression "a"+"b" is evaluated at runtime to the string "ab".
- The key "ab" exists in the dictionary, so lookup returns 7 and prints it. No error.

3) Intermediate evaluation (for Option B)
- dict created: {"a":3,"b":2,"c":7}
- Evaluate dict[a]: attempt to evaluate a -> NameError: name 'a' is not defined

4) Why the correct answer is correct
Option B attempts to use an undefined variable a as the key expression. That triggers a NameError when Python evaluates the expression inside the brackets, causing a runtime error after the dictionary is created.

5) Why other options are incorrect
- A only prints the dict (no error).
- C uses the string literal "a" (valid lookup).
- D concatenates string literals to "ab", which exists as a key (valid lookup).

6) Conclusion
Therefore the line that will run but then give a runtime error is: dict = {"a": 3, "b": 2, "c":7}
print(dict[a]) — this matches the provided correct answer.

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