No implicit fallthrough in switch in Swift - Step-by-Step Execution

Concept Flow - No implicit fallthrough in switch

Start switch with value

↓

Match case 1?

No→Match case 2?

↓

Execute case 1

↓

Exit switch

Swift checks each case one by one and runs only the matching case's code without automatically running the next cases.

Execution Sample

Swift

let number = 2
switch number {
case 1:
    print("One")
case 2:
    print("Two")
default:
    print("Other")
}

This code prints the word for the number 2 using a switch statement without running other cases.

Execution Table

Step	Condition Checked	Result	Action	Output
1	number == 1	False	Skip case 1
2	number == 2	True	Execute case 2	Two
3	Exit switch	N/A	Stop checking further cases

💡 Case 2 matched, so switch exits without running other cases.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
number	2	2	2	2

Key Moments - 2 Insights

Why doesn't the switch run the code in case 3 after case 2 matches?

Swift switch statements do not have implicit fallthrough, so after a matching case runs (case 2), the switch exits immediately (see execution_table step 3).

What happens if no case matches the value?

The default case runs if no other case matches, ensuring some code always executes (not shown in this example but explained in concept_flow).

Visual Quiz - 3 Questions

Look at the execution table, what is the output at step 2?

A "One"

B "Two"

C "Other"

D No output

💡 Hint

Check the 'Output' column in execution_table row for step 2.

At which step does the switch stop checking cases?

A Step 3

B Step 2

C Step 1

D Never stops

💡 Hint

Look at the 'Action' column in execution_table step 3.

If number was 3, which case would run?

A case 1

B case 2

C default

D No case runs

💡 Hint

Recall the default case runs when no other case matches, as described in concept_flow.

Concept Snapshot

Swift switch checks each case for a match.
Only the matching case runs; no automatic fallthrough.
Use 'default' to catch unmatched values.
No need for break statements.
Switch exits after running matched case.

Full Transcript

This visual trace shows how Swift's switch statement works without implicit fallthrough. The switch checks each case in order. When it finds a matching case, it runs that case's code and then exits immediately without running other cases. For example, when number is 2, case 2 matches, so it prints "Two" and stops. If no case matches, the default case runs. This behavior means you don't need break statements to stop execution like in some other languages.

Practice

1. What happens if you do not use fallthrough in a Swift switch statement?

easy

A. Only the matched case runs; no automatic fallthrough to the next case.

B. All cases after the matched one run automatically.

C. The program throws a runtime error.

D. The switch statement ignores the matched case.

Solution

1. Step 1: Understand Swift switch behavior

   Swift switch statements do not automatically continue to the next case after a match.

2. Step 2: Role of fallthrough keyword

   To continue to the next case, you must explicitly use the fallthrough keyword.

3. Final Answer:

   Only the matched case runs; no automatic fallthrough to the next case. -> Option A

4. Quick Check:

   No implicit fallthrough = matched case only [OK]

Hint: Remember: Swift switch stops after matched case unless fallthrough used [OK]

Common Mistakes:

• Assuming all cases run after a match like in C
• Forgetting to add fallthrough when needed
• Expecting runtime errors without fallthrough
• Thinking switch ignores matched cases

2. Which of the following is the correct syntax to explicitly continue to the next case in a Swift switch?

easy

A. Use the keyword continue inside the case.

B. No syntax needed; Swift does it automatically.

C. Use the keyword fallthrough inside the case.

D. Use a comma , between cases.

Solution

1. Step 1: Identify the keyword for fallthrough

   Swift uses the keyword fallthrough to continue to the next case.

2. Step 2: Eliminate incorrect options

   continue is for loops, commas separate cases but don't cause fallthrough, and Swift does not do it automatically.

3. Final Answer:

   Use the keyword fallthrough inside the case. -> Option C

4. Quick Check:

   Explicit fallthrough keyword = fallthrough [OK]

Hint: Use fallthrough keyword to continue to next case in Swift switch [OK]

Common Mistakes:

• Confusing fallthrough with continue
• Using commas to join cases expecting fallthrough
• Assuming automatic fallthrough without keyword
• Using break to continue to next case

3. What is the output of the following Swift code?

let number = 2
switch number {
 case 1:
   print("One")
 case 2:
   print("Two")
   fallthrough
 case 3:
   print("Three")
 default:
   print("Other")
}

medium

A. Three

B. Two
Three

C. Two

D. Two
Three
Other

Solution

1. Step 1: Identify matched case

   The value 2 matches case 2, so it prints "Two".

2. Step 2: Check for fallthrough

   Since fallthrough is used, execution continues to case 3, printing "Three".

3. Final Answer:

   Two Three -> Option B

4. Quick Check:

   fallthrough causes next case to run = Two and Three [OK]

Hint: Look for fallthrough keyword to know if next case runs [OK]

Common Mistakes:

• Ignoring fallthrough and printing only "Two"
• Printing "Three" only without "Two"
• Expecting default case to run
• Assuming all cases print

4. Identify the error in this Swift switch statement:

let value = 3
switch value {
 case 1:
   print("One")
 case 2:
   print("Two")
 case 3:
   print("Three")
   fallthrough
 case 4:
   print("Four")
   fallthrough
 case 5:
   print("Five")
}

medium

A. Missing default case causes a compile error.

B. Multiple fallthroughs cause a runtime error.

C. No error; code runs and prints Three, Four, Five.

D. Fallthrough cannot be used in the last case.

Solution

1. Step 1: Exhaustiveness requirement

   Swift switch on Int must be exhaustive: cover all possible values or add default. Cases 1-5 don't cover all Ints, no default: compile error.

2. Step 2: Fallthrough usage is valid

   Fallthroughs in cases 3 and 4 are allowed; no fallthrough in case 5 (correct); multiples OK, no runtime issues.

3. Final Answer:

   Missing default case causes a compile error. -> Option A

4. Quick Check:

   Non-exhaustive switch on Int = compile error [OK]

Hint: Swift switch on Int needs default unless all cases covered [OK]

Common Mistakes:

• Picking A: no fallthrough in last case
• Picking B: multiple fallthroughs allowed
• Picking C: ignores compile-time exhaustiveness error
• Thinking value=3 makes it exhaustive

5. You want to print all numbers from a matched case up to 5 in a Swift switch. Which code correctly uses fallthrough to achieve this?

let num = 3
switch num {
 case 1:
   print(1)
   fallthrough
 case 2:
   print(2)
   fallthrough
 case 3:
   print(3)
   fallthrough
 case 4:
   print(4)
   fallthrough
 case 5:
   print(5)
 default:
   print("Done")
}

hard

A. This code prints 1 to 5 regardless of num.

B. This code prints only 3 when num is 3.

C. This code causes a compile error due to fallthrough in default.

D. This code prints 3, 4, 5 when num is 3.

Solution

1. Step 1: Identify matched case and fallthrough chain

   When num is 3, case 3 matches and prints 3, then fallthrough continues to cases 4 and 5 printing 4 and 5.

2. Step 2: Confirm output and default usage

   Default is not reached because cases 3 to 5 cover the fallthrough chain; no error occurs.

3. Final Answer:

   This code prints 3, 4, 5 when num is 3. -> Option D

4. Quick Check:

   Fallthrough chains print from matched case to end [OK]

Hint: Use fallthrough in each case to print all following cases [OK]

Common Mistakes:

• Expecting only matched case to print
• Thinking default runs after fallthrough chain
• Assuming compile error from fallthrough in default
• Believing all cases print regardless of match