Break switch ошибка

Here’s an instance of implementing
break statement
s in a switch statement.

let numberSymbol: Character = "三"  // Simplified Chinese for the number 3
var possibleIntegerValue: Int?
switch numberSymbol {
case "1", "١", "一", "๑":
    possibleIntegerValue = 1
case "2", "٢", "二", "๒":
    possibleIntegerValue = 2
case "3", "٣", "三", "๓":
    possibleIntegerValue = 3
case "4", "٤", "四", "๔":
    possibleIntegerValue = 4
default:
    break
}
if let integerValue = possibleIntegerValue {
    println("The integer value of \(numberSymbol) is \(integerValue).")
} else {
    println("An integer value could not be found for \(numberSymbol).")
}

The utilization of breaks in switch statement (

possibleIntegerValue

) is not necessarily superior to other options. An alternative such as (

possibleIntegerValue = nil

) can be just as effective. This is due to the fact that (

optional Int

).

let numberSymbol: Character = "三"  // Simplified Chinese for the number 3
var possibleIntegerValue: Int?
switch numberSymbol {
case "1", "١", "一", "๑":
    possibleIntegerValue = 1
case "2", "٢", "二", "๒":
    possibleIntegerValue = 2
case "3", "٣", "三", "๓":
    possibleIntegerValue = 3
case "4", "٤", "四", "๔":
    possibleIntegerValue = 4
default:
    possibleIntegerValue = nil
}
if let integerValue = possibleIntegerValue {
    println("The integer value of \(numberSymbol) is \(integerValue).")
} else {
    println("An integer value could not be found for \(numberSymbol).")
}

In this scenario, there is no need for

break

. Could someone provide me with an appropriate illustration of utilizing breaks in a switch statement, where certain cases are intentionally disregarded?

The book says:

To match and ignore cases in a
switch statement
, you can use this technique. The
Swift’s switch
statement is exhaustive and doesn’t permit empty cases. However, there may be cases where you deliberately want to match and disregard a case to make your intentions clear. In such cases, you can write the break statement as the entire body of the case you want to ignore, and this will cause the switch statement’s execution to end immediately when that specific case is matched.

When you need to include a case in your logic for clarity purposes, but it does not require any actual functionality, you can use the

break

statement inside a

switch

. For instance, let’s say you want to use a

switch

statement to determine whether to
Determine if a given year is a leap year
or not, even though it may seem like a contrived example.

func isLeapYear(year: Int) -> Bool {
    switch (year) {
    case let x where (x % 100) == 0 && (x % 400) != 0:
        break
    case let x where (x % 4) == 0:
        return true
    default:
        break
    }
    return false
}
isLeapYear(2014)    // false
isLeapYear(2000)    // true
isLeapYear(1900)    // false

In

isLeapYear

, utilizing the first instance of

switch

helps in identifying the exceptional non-leap years, which are the years that are both divisible by 100 and not divisible by 400. In such cases, the

break

statement corresponds to taking no action.

Here’s the issue at hand: I’m working with a switch-statement that requires me to verify if a particular string matches a specific symbol. However, the current code implementation seems to be causing some trouble.

[...]
switch($str) { 
    [...]
    case (: $output .= $example_array[rand(0,3)] . '|'; break;
    case ): $output .= $example_array[rand(4,8)] . '|'; break;
    [...]
}
[...]

I get the following error:

An error occurred while parsing the code, specifically an unexpected ‘case’ (T_CASE) in […]

Is there a way to avoid changing my statement to either

if

or

else

, or can I find a workaround?

It could be helpful to review the structure of the code in

switch

.

To match the desired character, it is necessary to enclose it in quotation marks.

case "(":
  $output .= $example_array[rand(0,3)] . '|';
  break;

When assessing a string that contains parentheses, it is recommended to use quotation marks.

case ')':
case '(':

I am perplexed as to where I am going wrong in my code. My objective is to accomplish the following task:

switch (action.type) {
    case TYPES.ADD_TO_FAVORITES:
      if (state.faves.length <= 10) {
        return assign({}, state, {
          faves: state.faves.concat(action.payload),
          full: false
        });
      } else {
        return assign({}, state, {
          faves: state.faves,
          full: true
        });
      }
    default:
      return state;
  }

According to my code checker, I need to include a break prior to the case labelled

default

. However, upon implementing this suggestion, my checker then gives me an error message denoted by

unreachable code

.

The eslint linter rule ‘no-fallthrough’ is designed to prevent inadvertent fallthrough from one case to another.

If there is no break or return encountered, the code execution will persist from the matching case to the following cases.

At times, there may be a requirement for this, but it is essential to avoid unintentional fallthrough, which is precisely what this rule aims to prevent.

Disabling the rule or setting it as a warning can be done. However, it is suggested to assign a variable for the return value within the function and to simply return it at the end, rather than disabling the rule altogether.

function() {
var returnvalue;
Switch(variableA) {
    Case 1:
        returnvalue = somevalue;
        break;
    case 2:
        returnvalue = some other value;
        break;
     default:
         returnvalue= default value;
    }
return returnvalue; 

}

Upon exiting your if-else block, you will be returning from the portion that cannot be reached.

The break statement will not have an opportunity to be executed.

switch (action.type) {
case TYPES.ADD_TO_FAVORITES:
  if (state.faves.length <= 10) {
    return ...;
  } else {
    return ...;
  }
  break; // YOU HAVE ADDED BREAK HERE
default:
  return state;
}

In

case TYPES.ADD_TO_FAVORITES

, either

if

or

else

will be executed. If an object is returned in either the if or

else

, then any code added after

break

and before

default

will never be executed.

The reason for the inclusion of

unreachable code

is explicitly stated.

It is not sensible and prohibited to have a return statement within a switch block. The only valid place to use the return statement is inside a function. To obtain a value from the switch statement, you may assign it to a variable.

var result;
switch (action.type) {
    case TYPES.ADD_TO_FAVORITES:
        if (state.faves.length <= 10) {
            result = assign({}, state, {
                faves: state.faves.concat(action.payload),
                full: false
            });
        } else {
            result = assign({}, state, {
                faves: state.faves,
                full: true
            });
        }
        break;
    default:
        result = state;
        break;
}

Presuming the code example (taken from the source):

#include 
void playgame()
{
    printf( "Play game called" );
}
void loadgame()
{
    printf( "Load game called" );
}
void playmultiplayer()
{
    printf( "Play multiplayer game called" );
}
int main()
{
    int input;
    printf( "1. Play game\n" );
    printf( "2. Load game\n" );
    printf( "3. Play multiplayer\n" );
    printf( "4. Exit\n" );
    printf( "Selection: " );
    scanf( "%d", &input );
    switch ( input ) {
        case 1:            /* Note the colon, not a semicolon */
            playgame();
            break;
        case 2:
            loadgame();
            break;
        case 3:
            playmultiplayer();
            break;
        case 4:
            printf( "Thanks for playing!\n" );
            break;
        default:
            printf( "Bad input, quitting!\n" );
            break;
    }
    getchar();
    return 0;
}

Is it necessary to utilize a

break;

in the final

default

scenario? Although the program functions in the same way when it is removed, I have observed that other instances implement a

break;

in the

default

situation.

Why? Is there a reason?

Is it necessary to include a break statement in the final default case?

Here’s an excerpt from the second edition of «The C programming language,» also known as K&R 2.

Section 3.4 focuses on the topic of switches.

It’s always good practice to include a break statement after the final case, even if it’s not logically required. This defensive programming technique can prove to be helpful in the future, in case a new case is added at the end.

Consider the rationale behind implementing

break

within a switch statement. Take into account the scenario without
breaking switch statement
.

switch ( input ) {
    case 1:            /* Note the colon, not a semicolon */
        playgame();
    case 2:
        loadgame();
    case 3:
        playmultiplayer();
    case 4:
        printf( "Thanks for playing!\n" );
    default:
        printf( "Bad input, quitting!\n" );
}

If

input == 1

is supposed to happen, the program will naturally invoke

playgame()

. However, if there’s a lack of

break

, the program cannot complete the

switch

and will instead execute

loadgame()

, followed by

playmultiplayer()

, and two

printf

s consecutively.

In order to prevent it, we employ

break

.

case 1:
    playgame();
    break; /* here */
case 2:
    ...

The expected outcome is for the program to finish executing the

switch

statement before moving on to the codes of

case 2

, which is due to

break

.

Your switch is this:

switch ( input ) {
    case 1:            /* Note the colon, not a semicolon */
        playgame();
        break;
    case 2:
        loadgame();
        break;
    case 3:
        playmultiplayer();
        break;
    case 4:
        printf( "Thanks for playing!\n" );
        break;
    default:
        printf( "Bad input, quitting!\n" );
        break;
}

If there is no occurrence of

case

following

default

, then it makes no difference whether or not you include

break

on

default

. Nevertheless, it is simple to assume a new scenario to write.

    default:
        printf( "Thanks for playing!\n" );
        /* what happens if there's no `break`...? */
    case 5:
        the_new_feature();
        break;
}

It is typical to make errors in C/C++, and if a new feature is added and then forgotten after five years, it can result in a highly flawed bug. In fact, some contemporary languages like C# prohibit switch-case without either

break

or

return

.

The syntax appears to be error-free; however, it is strongly advised to avoid the practice mentioned and instead opt for

break

.

The outcome is contingent on the manner in which the default case is composed.

In this particular instance, it is essential to have a break.

switch ( input ) {
    default:
        printf( "Bad input, quitting!\n" );
        break;
    case 1:            /* Note the colon, not a semicolon */
        playgame();
        break;
    case 2:
        loadgame();
        break;
}

Whether you exclude or retain the

break

statement, it is merely a matter of preference and will not alter the outcome.

It is beneficial to include the break statement in code for various purposes.

Have a glance at the programmers’ forum.