Why does the code print an optional value?

-1

var myString = "7"

var possibleInt = Int(myString)

print(possibleInt)

The above code prints out:

Optional(7)

Why does it do this even though I did not use an optional. The string was converted to a number so shouldn't the out be:

7

myString = "banana"

possibleInt = Int(myString)

print(possibleInt)

This code will print out nil because type conversion did not work but I don't see how 7 could be an optional.

asked Sep 27 '15 at 0:15

Nocturnal

16010

1

If it can be nil, it has to be an Optional.

– matt
Sep 27 '15 at 0:17

@matt Why can it be a nil?

– Nocturnal
Sep 27 '15 at 0:18

1

As you yourself said. It might fail because the string is not a number.

– matt
Sep 27 '15 at 0:19

1

The Int(string:) constructor can only return 1 type. It doesn't get to decide that it is going to return a regular Int if the conversion succeeds and a nil if it doesn't. It is not possible to return both nil and an Int. If you're going to return nil, you have to return an Int? which is what Int(string:) returns. The return type of a function is determined at compile time, but the value you are passing to be be converted can be determined at run time.

– vacawama
Sep 27 '15 at 0:29

1

If you don't want it to be an optional you have to provide a default value. You can use ?? nil coalescing operator. let num = Int("string") ?? 0

– Leo Dabus
Sep 27 '15 at 0:31

|
show 1 more comment

-1

var myString = "7"

var possibleInt = Int(myString)

print(possibleInt)

The above code prints out:

Optional(7)

Why does it do this even though I did not use an optional. The string was converted to a number so shouldn't the out be:

7

myString = "banana"

possibleInt = Int(myString)

print(possibleInt)

This code will print out nil because type conversion did not work but I don't see how 7 could be an optional.

asked Sep 27 '15 at 0:15

Nocturnal

16010

1

If it can be nil, it has to be an Optional.

– matt
Sep 27 '15 at 0:17

@matt Why can it be a nil?

– Nocturnal
Sep 27 '15 at 0:18

1

As you yourself said. It might fail because the string is not a number.

– matt
Sep 27 '15 at 0:19

1

The Int(string:) constructor can only return 1 type. It doesn't get to decide that it is going to return a regular Int if the conversion succeeds and a nil if it doesn't. It is not possible to return both nil and an Int. If you're going to return nil, you have to return an Int? which is what Int(string:) returns. The return type of a function is determined at compile time, but the value you are passing to be be converted can be determined at run time.

– vacawama
Sep 27 '15 at 0:29

1

If you don't want it to be an optional you have to provide a default value. You can use ?? nil coalescing operator. let num = Int("string") ?? 0

– Leo Dabus
Sep 27 '15 at 0:31

|
show 1 more comment

-1

var myString = "7"

var possibleInt = Int(myString)

print(possibleInt)

The above code prints out:

Optional(7)

Why does it do this even though I did not use an optional. The string was converted to a number so shouldn't the out be:

7

myString = "banana"

possibleInt = Int(myString)

print(possibleInt)

This code will print out nil because type conversion did not work but I don't see how 7 could be an optional.

asked Sep 27 '15 at 0:15

Nocturnal

16010

var myString = "7"

var possibleInt = Int(myString)

print(possibleInt)

The above code prints out:

Optional(7)

Why does it do this even though I did not use an optional. The string was converted to a number so shouldn't the out be:

7

myString = "banana"

possibleInt = Int(myString)

print(possibleInt)

This code will print out nil because type conversion did not work but I don't see how 7 could be an optional.

xcode swift optional

asked Sep 27 '15 at 0:15

Nocturnal

16010

asked Sep 27 '15 at 0:15

Nocturnal

16010

asked Sep 27 '15 at 0:15

Nocturnal

16010

asked Sep 27 '15 at 0:15

Nocturnal

16010

asked Sep 27 '15 at 0:15

Nocturnal

16010

1

If it can be nil, it has to be an Optional.

– matt
Sep 27 '15 at 0:17

@matt Why can it be a nil?

– Nocturnal
Sep 27 '15 at 0:18

1

As you yourself said. It might fail because the string is not a number.

– matt
Sep 27 '15 at 0:19

1

The Int(string:) constructor can only return 1 type. It doesn't get to decide that it is going to return a regular Int if the conversion succeeds and a nil if it doesn't. It is not possible to return both nil and an Int. If you're going to return nil, you have to return an Int? which is what Int(string:) returns. The return type of a function is determined at compile time, but the value you are passing to be be converted can be determined at run time.

– vacawama
Sep 27 '15 at 0:29

1

If you don't want it to be an optional you have to provide a default value. You can use ?? nil coalescing operator. let num = Int("string") ?? 0

– Leo Dabus
Sep 27 '15 at 0:31

|
show 1 more comment

1

If it can be nil, it has to be an Optional.

– matt
Sep 27 '15 at 0:17

@matt Why can it be a nil?

– Nocturnal
Sep 27 '15 at 0:18

1

As you yourself said. It might fail because the string is not a number.

– matt
Sep 27 '15 at 0:19

1

The Int(string:) constructor can only return 1 type. It doesn't get to decide that it is going to return a regular Int if the conversion succeeds and a nil if it doesn't. It is not possible to return both nil and an Int. If you're going to return nil, you have to return an Int? which is what Int(string:) returns. The return type of a function is determined at compile time, but the value you are passing to be be converted can be determined at run time.

– vacawama
Sep 27 '15 at 0:29

1

If you don't want it to be an optional you have to provide a default value. You can use ?? nil coalescing operator. let num = Int("string") ?? 0

– Leo Dabus
Sep 27 '15 at 0:31

If it can be nil, it has to be an Optional.

– matt
Sep 27 '15 at 0:17

@matt Why can it be a nil?

– Nocturnal
Sep 27 '15 at 0:18

As you yourself said. It might fail because the string is not a number.

– matt
Sep 27 '15 at 0:19

The Int(string:) constructor can only return 1 type. It doesn't get to decide that it is going to return a regular Int if the conversion succeeds and a nil if it doesn't. It is not possible to return both nil and an Int. If you're going to return nil, you have to return an Int? which is what Int(string:) returns. The return type of a function is determined at compile time, but the value you are passing to be be converted can be determined at run time.

– vacawama
Sep 27 '15 at 0:29

If you don't want it to be an optional you have to provide a default value. You can use ?? nil coalescing operator. let num = Int("string") ?? 0

– Leo Dabus
Sep 27 '15 at 0:31

|
show 1 more comment

4 Answers
4

active

oldest

votes

When you convert a value (i.e. a String to an Int) it will be an optional - if you tried to run:

let num = Int("things")

// num is Optional(nil)

This couldn't possibly be a number, so it returns an 'empty' optional.

If you know that the string will always be a valid number you can unwrap it unsafely:

let num = Int("5")!

// num is 5, not Optional(5)

Or if you don't know if the string will be a valid int, you can safely unwrap it:

if let num = Int(myString) {

     // Do something with the number

} else {

    // Catch some error

}

edited Sep 27 '15 at 19:23

answered Sep 27 '15 at 0:19

Will Richardson

4,95953343

1

The catch with that is that if you do let num = Int("Hello")! then the app will crash at runtime. It would be better to add some optional binding rather than explicit unwrapping.

– Fogmeister
Sep 27 '15 at 15:18

add a comment |

The beauty about optionals is that they create a safety net in your code. If you take another language and try to parse "example123" as an Int, it will crash and the program will terminate. However in Swift, most functions which may otherwise lead to a crash return an optional, so that the program keeps running.

There are two ways you can avoid getting an optional value at the end of your program.

If you are absolutely sure that the input in Int() will be an integer, you can use

var possibleInt = Int(myString)!

And of course, if myString isn't an integer, your app will crash. However you can securely unwrap it by doing

if let possibleInt = Int(myString) {

    // use possibleInt as an int and not an optional

}

// otherwise, the program won't do anything with possibleInt

Hope this helps.

answered Sep 27 '15 at 15:08

user5381742

add a comment |

Int() method return a optional value, because if your string contain something that is not a number, such as "abc123", Int("abc123") will return nil,
and also, a method return a optional value has not any relationship with what is the method's argument is.

answered Sep 27 '15 at 2:35

A BLUE

184

add a comment |

You can see you did not declare a "myString" to any type first declare as string/integer and then try to print that variable. you will get the exact result that you want.

var myString = "7"

var possibleInt: Int;

var possibleInt = Int(myString)

print(possibleInt)

edited Nov 15 '18 at 5:21

answered Nov 14 '18 at 12:44

Tatsat

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

When you convert a value (i.e. a String to an Int) it will be an optional - if you tried to run:

let num = Int("things")

// num is Optional(nil)

This couldn't possibly be a number, so it returns an 'empty' optional.

If you know that the string will always be a valid number you can unwrap it unsafely:

let num = Int("5")!

// num is 5, not Optional(5)

Or if you don't know if the string will be a valid int, you can safely unwrap it:

if let num = Int(myString) {

     // Do something with the number

} else {

    // Catch some error

}

edited Sep 27 '15 at 19:23

answered Sep 27 '15 at 0:19

Will Richardson

4,95953343

1

The catch with that is that if you do let num = Int("Hello")! then the app will crash at runtime. It would be better to add some optional binding rather than explicit unwrapping.

– Fogmeister
Sep 27 '15 at 15:18

add a comment |

When you convert a value (i.e. a String to an Int) it will be an optional - if you tried to run:

let num = Int("things")

// num is Optional(nil)

This couldn't possibly be a number, so it returns an 'empty' optional.

If you know that the string will always be a valid number you can unwrap it unsafely:

let num = Int("5")!

// num is 5, not Optional(5)

Or if you don't know if the string will be a valid int, you can safely unwrap it:

if let num = Int(myString) {

     // Do something with the number

} else {

    // Catch some error

}

edited Sep 27 '15 at 19:23

answered Sep 27 '15 at 0:19

Will Richardson

4,95953343

1

The catch with that is that if you do let num = Int("Hello")! then the app will crash at runtime. It would be better to add some optional binding rather than explicit unwrapping.

– Fogmeister
Sep 27 '15 at 15:18

add a comment |

When you convert a value (i.e. a String to an Int) it will be an optional - if you tried to run:

let num = Int("things")

// num is Optional(nil)

This couldn't possibly be a number, so it returns an 'empty' optional.

If you know that the string will always be a valid number you can unwrap it unsafely:

let num = Int("5")!

// num is 5, not Optional(5)

Or if you don't know if the string will be a valid int, you can safely unwrap it:

if let num = Int(myString) {

     // Do something with the number

} else {

    // Catch some error

}

edited Sep 27 '15 at 19:23

answered Sep 27 '15 at 0:19

Will Richardson

4,95953343

When you convert a value (i.e. a String to an Int) it will be an optional - if you tried to run:

let num = Int("things")

// num is Optional(nil)

This couldn't possibly be a number, so it returns an 'empty' optional.

If you know that the string will always be a valid number you can unwrap it unsafely:

let num = Int("5")!

// num is 5, not Optional(5)

Or if you don't know if the string will be a valid int, you can safely unwrap it:

if let num = Int(myString) {

     // Do something with the number

} else {

    // Catch some error

}

edited Sep 27 '15 at 19:23

answered Sep 27 '15 at 0:19

Will Richardson

4,95953343

edited Sep 27 '15 at 19:23

answered Sep 27 '15 at 0:19

Will Richardson

4,95953343

answered Sep 27 '15 at 0:19

Will Richardson

4,95953343

answered Sep 27 '15 at 0:19

Will Richardson

4,95953343

1

The catch with that is that if you do let num = Int("Hello")! then the app will crash at runtime. It would be better to add some optional binding rather than explicit unwrapping.

– Fogmeister
Sep 27 '15 at 15:18

add a comment |

1

The catch with that is that if you do let num = Int("Hello")! then the app will crash at runtime. It would be better to add some optional binding rather than explicit unwrapping.

– Fogmeister
Sep 27 '15 at 15:18

The catch with that is that if you do let num = Int("Hello")! then the app will crash at runtime. It would be better to add some optional binding rather than explicit unwrapping.

– Fogmeister
Sep 27 '15 at 15:18

add a comment |

There are two ways you can avoid getting an optional value at the end of your program.

If you are absolutely sure that the input in Int() will be an integer, you can use

var possibleInt = Int(myString)!

And of course, if myString isn't an integer, your app will crash. However you can securely unwrap it by doing

if let possibleInt = Int(myString) {

    // use possibleInt as an int and not an optional

}

// otherwise, the program won't do anything with possibleInt

Hope this helps.

answered Sep 27 '15 at 15:08

user5381742

add a comment |

There are two ways you can avoid getting an optional value at the end of your program.

If you are absolutely sure that the input in Int() will be an integer, you can use

var possibleInt = Int(myString)!

And of course, if myString isn't an integer, your app will crash. However you can securely unwrap it by doing

if let possibleInt = Int(myString) {

    // use possibleInt as an int and not an optional

}

// otherwise, the program won't do anything with possibleInt

Hope this helps.

answered Sep 27 '15 at 15:08

user5381742

add a comment |

There are two ways you can avoid getting an optional value at the end of your program.

If you are absolutely sure that the input in Int() will be an integer, you can use

var possibleInt = Int(myString)!

And of course, if myString isn't an integer, your app will crash. However you can securely unwrap it by doing

if let possibleInt = Int(myString) {

    // use possibleInt as an int and not an optional

}

// otherwise, the program won't do anything with possibleInt

Hope this helps.

answered Sep 27 '15 at 15:08

user5381742

There are two ways you can avoid getting an optional value at the end of your program.

If you are absolutely sure that the input in Int() will be an integer, you can use

var possibleInt = Int(myString)!

And of course, if myString isn't an integer, your app will crash. However you can securely unwrap it by doing

if let possibleInt = Int(myString) {

    // use possibleInt as an int and not an optional

}

// otherwise, the program won't do anything with possibleInt

Hope this helps.

answered Sep 27 '15 at 15:08

user5381742

answered Sep 27 '15 at 15:08

user5381742

answered Sep 27 '15 at 15:08

user5381742

answered Sep 27 '15 at 15:08

user5381742

add a comment |

answered Sep 27 '15 at 2:35

A BLUE

184

add a comment |

answered Sep 27 '15 at 2:35

A BLUE

184

add a comment |

answered Sep 27 '15 at 2:35

A BLUE

184

answered Sep 27 '15 at 2:35

A BLUE

184

answered Sep 27 '15 at 2:35

A BLUE

184

answered Sep 27 '15 at 2:35

A BLUE

184

answered Sep 27 '15 at 2:35

A BLUE

184

add a comment |

You can see you did not declare a "myString" to any type first declare as string/integer and then try to print that variable. you will get the exact result that you want.

var myString = "7"

var possibleInt: Int;

var possibleInt = Int(myString)

print(possibleInt)

edited Nov 15 '18 at 5:21

answered Nov 14 '18 at 12:44

Tatsat

add a comment |

You can see you did not declare a "myString" to any type first declare as string/integer and then try to print that variable. you will get the exact result that you want.

var myString = "7"

var possibleInt: Int;

var possibleInt = Int(myString)

print(possibleInt)

edited Nov 15 '18 at 5:21

answered Nov 14 '18 at 12:44

Tatsat

add a comment |

You can see you did not declare a "myString" to any type first declare as string/integer and then try to print that variable. you will get the exact result that you want.

var myString = "7"

var possibleInt: Int;

var possibleInt = Int(myString)

print(possibleInt)

edited Nov 15 '18 at 5:21

answered Nov 14 '18 at 12:44

Tatsat

You can see you did not declare a "myString" to any type first declare as string/integer and then try to print that variable. you will get the exact result that you want.

var myString = "7"

var possibleInt: Int;

var possibleInt = Int(myString)

print(possibleInt)

edited Nov 15 '18 at 5:21

answered Nov 14 '18 at 12:44

Tatsat

edited Nov 15 '18 at 5:21

answered Nov 14 '18 at 12:44

Tatsat

answered Nov 14 '18 at 12:44

Tatsat

answered Nov 14 '18 at 12:44

Tatsat

add a comment |

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