Says type is non-optional but prints optional

up vote
-2
down vote

favorite

I have this code:

let result_string = "(String(describing: value))"

because I need to convert this from type Any to String. However when I print result string, it prints:

optional(abcd)

How do I make it print just

abcd

When I tried the other fixes for optionals it told me that result_string is not optional and didn't let me force unwrap.

EDIT:
Value was an optional, so force unwrapping it fixed my problem.

let result_string = "(String(describing: value!))"

edited yesterday

asked yesterday

Athreya Daniel

New contributor

My guess: value is an optional ...
– Martin R
yesterday

@Martin R So should I try force unwrapping value?
– Athreya Daniel
yesterday

@MartinR Thank You, force unwrapping this fixed my problem.
– Athreya Daniel
yesterday

2

force unwrap may crash if the value is nil
– Sh_Khan
yesterday

@AthreyaDaniel your result_string is optional but value which is type Any is not optional and you are trying to force unwrap non-optional value.
– Prabhat
10 hours ago

|
show 1 more comment

up vote
-2
down vote

favorite

I have this code:

let result_string = "(String(describing: value))"

because I need to convert this from type Any to String. However when I print result string, it prints:

optional(abcd)

How do I make it print just

abcd

When I tried the other fixes for optionals it told me that result_string is not optional and didn't let me force unwrap.

EDIT:
Value was an optional, so force unwrapping it fixed my problem.

let result_string = "(String(describing: value!))"

edited yesterday

asked yesterday

Athreya Daniel

New contributor

My guess: value is an optional ...
– Martin R
yesterday

@Martin R So should I try force unwrapping value?
– Athreya Daniel
yesterday

@MartinR Thank You, force unwrapping this fixed my problem.
– Athreya Daniel
yesterday

2

force unwrap may crash if the value is nil
– Sh_Khan
yesterday

@AthreyaDaniel your result_string is optional but value which is type Any is not optional and you are trying to force unwrap non-optional value.
– Prabhat
10 hours ago

|
show 1 more comment

up vote
-2
down vote

favorite

I have this code:

let result_string = "(String(describing: value))"

because I need to convert this from type Any to String. However when I print result string, it prints:

optional(abcd)

How do I make it print just

abcd

When I tried the other fixes for optionals it told me that result_string is not optional and didn't let me force unwrap.

EDIT:
Value was an optional, so force unwrapping it fixed my problem.

let result_string = "(String(describing: value!))"

edited yesterday

asked yesterday

Athreya Daniel

New contributor

I have this code:

let result_string = "(String(describing: value))"

because I need to convert this from type Any to String. However when I print result string, it prints:

optional(abcd)

How do I make it print just

abcd

When I tried the other fixes for optionals it told me that result_string is not optional and didn't let me force unwrap.

EDIT:
Value was an optional, so force unwrapping it fixed my problem.

let result_string = "(String(describing: value!))"

swift xcode

edited yesterday

asked yesterday

Athreya Daniel

New contributor

edited yesterday

asked yesterday

Athreya Daniel

New contributor

edited yesterday

asked yesterday

Athreya Daniel

New contributor

asked yesterday

Athreya Daniel

asked yesterday

Athreya Daniel

New contributor

Athreya Daniel is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

My guess: value is an optional ...
– Martin R
yesterday

@Martin R So should I try force unwrapping value?
– Athreya Daniel
yesterday

@MartinR Thank You, force unwrapping this fixed my problem.
– Athreya Daniel
yesterday

2

force unwrap may crash if the value is nil
– Sh_Khan
yesterday

@AthreyaDaniel your result_string is optional but value which is type Any is not optional and you are trying to force unwrap non-optional value.
– Prabhat
10 hours ago

|
show 1 more comment

My guess: value is an optional ...
– Martin R
yesterday

@Martin R So should I try force unwrapping value?
– Athreya Daniel
yesterday

@MartinR Thank You, force unwrapping this fixed my problem.
– Athreya Daniel
yesterday

2

force unwrap may crash if the value is nil
– Sh_Khan
yesterday

@AthreyaDaniel your result_string is optional but value which is type Any is not optional and you are trying to force unwrap non-optional value.
– Prabhat
10 hours ago

My guess: value is an optional ...
– Martin R
yesterday

@Martin R So should I try force unwrapping value?
– Athreya Daniel
yesterday

@MartinR Thank You, force unwrapping this fixed my problem.
– Athreya Daniel
yesterday

force unwrap may crash if the value is nil
– Sh_Khan
yesterday

@AthreyaDaniel your result_string is optional but value which is type Any is not optional and you are trying to force unwrap non-optional value.
– Prabhat
10 hours ago

|
show 1 more comment

2 Answers
2

active

oldest

votes

up vote
0
down vote

You can try

if let res = value {

  print(res)

}

guard let res = value { return }

print(res)

answered yesterday

Sh_Khan

34.2k41125

add a comment |

up vote
-1
down vote

you can try like below:

var value : Any? = "abcd"

let result_string = value as! String

print(result_string)

We have to downcast Any to string.
Worked in playground.

edited 8 hours ago

answered 10 hours ago

Prabhat

493

Sorry Prabhat but this is wrong. In your example, result_string is a non-optional string containing the literal string "Optional(value)". Same mistake as OP's.
– Moritz
9 hours ago

@Moritz If value is type Any with optional then it will produce Optional(value) and I was consi
– Prabhat
8 hours ago

Please test your code in a Playground, Prabhat. You will see that you get "initializer for conditional binding must have Optional type, not 'String'", because as I said, in your example result_string is not an optional. Hint: "interpolation"... ;)
– Moritz
8 hours ago

Thanks got it. We have to downcast value of type Any . I have edited the answer and verified it.
– Prabhat
8 hours ago

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
0
down vote

You can try

if let res = value {

  print(res)

}

guard let res = value { return }

print(res)

answered yesterday

Sh_Khan

34.2k41125

add a comment |

up vote
0
down vote

You can try

if let res = value {

  print(res)

}

guard let res = value { return }

print(res)

answered yesterday

Sh_Khan

34.2k41125

add a comment |

up vote
0
down vote

You can try

if let res = value {

  print(res)

}

guard let res = value { return }

print(res)

answered yesterday

Sh_Khan

34.2k41125

You can try

if let res = value {

  print(res)

}

guard let res = value { return }

print(res)

answered yesterday

Sh_Khan

34.2k41125

answered yesterday

Sh_Khan

34.2k41125

answered yesterday

Sh_Khan

34.2k41125

answered yesterday

Sh_Khan

34.2k41125

add a comment |

up vote
-1
down vote

you can try like below:

var value : Any? = "abcd"

let result_string = value as! String

print(result_string)

We have to downcast Any to string.
Worked in playground.

edited 8 hours ago

answered 10 hours ago

Prabhat

493

Sorry Prabhat but this is wrong. In your example, result_string is a non-optional string containing the literal string "Optional(value)". Same mistake as OP's.
– Moritz
9 hours ago

@Moritz If value is type Any with optional then it will produce Optional(value) and I was consi
– Prabhat
8 hours ago

Please test your code in a Playground, Prabhat. You will see that you get "initializer for conditional binding must have Optional type, not 'String'", because as I said, in your example result_string is not an optional. Hint: "interpolation"... ;)
– Moritz
8 hours ago

Thanks got it. We have to downcast value of type Any . I have edited the answer and verified it.
– Prabhat
8 hours ago

add a comment |

up vote
-1
down vote

you can try like below:

var value : Any? = "abcd"

let result_string = value as! String

print(result_string)

We have to downcast Any to string.
Worked in playground.

edited 8 hours ago

answered 10 hours ago

Prabhat

493

Sorry Prabhat but this is wrong. In your example, result_string is a non-optional string containing the literal string "Optional(value)". Same mistake as OP's.
– Moritz
9 hours ago

@Moritz If value is type Any with optional then it will produce Optional(value) and I was consi
– Prabhat
8 hours ago

Please test your code in a Playground, Prabhat. You will see that you get "initializer for conditional binding must have Optional type, not 'String'", because as I said, in your example result_string is not an optional. Hint: "interpolation"... ;)
– Moritz
8 hours ago

Thanks got it. We have to downcast value of type Any . I have edited the answer and verified it.
– Prabhat
8 hours ago

add a comment |

up vote
-1
down vote

you can try like below:

var value : Any? = "abcd"

let result_string = value as! String

print(result_string)

We have to downcast Any to string.
Worked in playground.

edited 8 hours ago

answered 10 hours ago

Prabhat

493

you can try like below:

var value : Any? = "abcd"

let result_string = value as! String

print(result_string)

We have to downcast Any to string.
Worked in playground.

edited 8 hours ago

answered 10 hours ago

Prabhat

493

edited 8 hours ago

answered 10 hours ago

Prabhat

493

answered 10 hours ago

Prabhat

493

answered 10 hours ago

Prabhat

493

Sorry Prabhat but this is wrong. In your example, result_string is a non-optional string containing the literal string "Optional(value)". Same mistake as OP's.
– Moritz
9 hours ago

@Moritz If value is type Any with optional then it will produce Optional(value) and I was consi
– Prabhat
8 hours ago

Please test your code in a Playground, Prabhat. You will see that you get "initializer for conditional binding must have Optional type, not 'String'", because as I said, in your example result_string is not an optional. Hint: "interpolation"... ;)
– Moritz
8 hours ago

Thanks got it. We have to downcast value of type Any . I have edited the answer and verified it.
– Prabhat
8 hours ago

add a comment |

Sorry Prabhat but this is wrong. In your example, result_string is a non-optional string containing the literal string "Optional(value)". Same mistake as OP's.
– Moritz
9 hours ago

@Moritz If value is type Any with optional then it will produce Optional(value) and I was consi
– Prabhat
8 hours ago

Please test your code in a Playground, Prabhat. You will see that you get "initializer for conditional binding must have Optional type, not 'String'", because as I said, in your example result_string is not an optional. Hint: "interpolation"... ;)
– Moritz
8 hours ago

Thanks got it. We have to downcast value of type Any . I have edited the answer and verified it.
– Prabhat
8 hours ago

Sorry Prabhat but this is wrong. In your example, result_string is a non-optional string containing the literal string "Optional(value)". Same mistake as OP's.
– Moritz
9 hours ago

@Moritz If value is type Any with optional then it will produce Optional(value) and I was consi
– Prabhat
8 hours ago

Please test your code in a Playground, Prabhat. You will see that you get "initializer for conditional binding must have Optional type, not 'String'", because as I said, in your example result_string is not an optional. Hint: "interpolation"... ;)
– Moritz
8 hours ago

Thanks got it. We have to downcast value of type Any . I have edited the answer and verified it.
– Prabhat
8 hours ago

add a comment |

Athreya Daniel is a new contributor. Be nice, and check out our Code of Conduct.

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