[Solved] Find the problem in this Swift code snippet:func outer() { func inner() { print(message) } let message = "Hello" inner() } outer() | Swift

Swift - Functions

Find the problem in this Swift code snippet:

func outer() {
  func inner() {
    print(message)
  }
  let message = "Hello"
  inner()
}
outer()

Aprint statement syntax is incorrect

Bmessage must be a global variable

Cinner() must be called before message declaration

Dinner() cannot access message declared after it

Step-by-Step Solution

Solution:

Step 1: Check variable scope and declaration order
message is declared after inner(), so inner() cannot access it because it is not yet defined.
Step 2: Understand Swift's variable visibility rules
Variables must be declared before use in the same scope or accessible via closure capture.
Final Answer:
inner() cannot access message declared after it -> Option D
Quick Check:
Variable must be declared before use in nested function [OK]

Quick Trick: Declare variables before nested functions that use them [OK]

Common Mistakes:

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