Avoiding string to be considered optional in swift












0















I have a function where string value is defined based on value in dictionary. If value in dictionary is not found, I set it to static string. String is constant (let): 



let dict = [ 1: "a", 

             2: "b" ]



func x() {

    let index = someFunctionReturnsInt()

    let str = (nil != dict[index]) ? dict[index] : "N/A"


So string cannot possibly be nil. However, when I try to use str, Swift still considers it an optional. For example:



    // still inside the same function

    os_log("Str is %@", str)


will return a compiler error:




Value of optional type 'String?' must be unwrapped to a value of type 'String'




Thing is, I use this function quite a few times, and adding str ?? "N/A" on every occasion makes no sense. Force-unwrap with str! seems like an overkill: this string is not critical to the app, so I don't think aborting execution if it's missing makes any sense.



So is force-unwrap the only solution, or is there a better way to define a string as non-optional and "never going to be nil"?






swift dictionary optional







share|improve this question




















  • 1





    You already know about the ?? operator. What is wrong with let str = dict[index] ?? "N/A"?

    – rob mayoff
    Nov 22 '18 at 17:51











  • @robmayoff you are right, didn't think it would make a difference

    – Kiril S.
    Nov 23 '18 at 19:25
















0















I have a function where string value is defined based on value in dictionary. If value in dictionary is not found, I set it to static string. String is constant (let): 



let dict = [ 1: "a", 

             2: "b" ]



func x() {

    let index = someFunctionReturnsInt()

    let str = (nil != dict[index]) ? dict[index] : "N/A"


So string cannot possibly be nil. However, when I try to use str, Swift still considers it an optional. For example:



    // still inside the same function

    os_log("Str is %@", str)


will return a compiler error:




Value of optional type 'String?' must be unwrapped to a value of type 'String'




Thing is, I use this function quite a few times, and adding str ?? "N/A" on every occasion makes no sense. Force-unwrap with str! seems like an overkill: this string is not critical to the app, so I don't think aborting execution if it's missing makes any sense.



So is force-unwrap the only solution, or is there a better way to define a string as non-optional and "never going to be nil"?






swift dictionary optional







share|improve this question




















  • 1





    You already know about the ?? operator. What is wrong with let str = dict[index] ?? "N/A"?

    – rob mayoff
    Nov 22 '18 at 17:51











  • @robmayoff you are right, didn't think it would make a difference

    – Kiril S.
    Nov 23 '18 at 19:25














0












0








0








I have a function where string value is defined based on value in dictionary. If value in dictionary is not found, I set it to static string. String is constant (let): 



let dict = [ 1: "a", 

             2: "b" ]



func x() {

    let index = someFunctionReturnsInt()

    let str = (nil != dict[index]) ? dict[index] : "N/A"


So string cannot possibly be nil. However, when I try to use str, Swift still considers it an optional. For example:



    // still inside the same function

    os_log("Str is %@", str)


will return a compiler error:




Value of optional type 'String?' must be unwrapped to a value of type 'String'




Thing is, I use this function quite a few times, and adding str ?? "N/A" on every occasion makes no sense. Force-unwrap with str! seems like an overkill: this string is not critical to the app, so I don't think aborting execution if it's missing makes any sense.



So is force-unwrap the only solution, or is there a better way to define a string as non-optional and "never going to be nil"?






swift dictionary optional







share|improve this question
















I have a function where string value is defined based on value in dictionary. If value in dictionary is not found, I set it to static string. String is constant (let): 



let dict = [ 1: "a", 

             2: "b" ]



func x() {

    let index = someFunctionReturnsInt()

    let str = (nil != dict[index]) ? dict[index] : "N/A"


So string cannot possibly be nil. However, when I try to use str, Swift still considers it an optional. For example:



    // still inside the same function

    os_log("Str is %@", str)


will return a compiler error:




Value of optional type 'String?' must be unwrapped to a value of type 'String'




Thing is, I use this function quite a few times, and adding str ?? "N/A" on every occasion makes no sense. Force-unwrap with str! seems like an overkill: this string is not critical to the app, so I don't think aborting execution if it's missing makes any sense.



So is force-unwrap the only solution, or is there a better way to define a string as non-optional and "never going to be nil"?






swift dictionary optional




swift dictionary optional






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 22 '18 at 19:01









Dávid Pásztor

22.6k83051




22.6k83051










asked Nov 22 '18 at 17:45









Kiril S.Kiril S.

5,73752245




5,73752245








  • 1





    You already know about the ?? operator. What is wrong with let str = dict[index] ?? "N/A"?

    – rob mayoff
    Nov 22 '18 at 17:51











  • @robmayoff you are right, didn't think it would make a difference

    – Kiril S.
    Nov 23 '18 at 19:25














  • 1





    You already know about the ?? operator. What is wrong with let str = dict[index] ?? "N/A"?

    – rob mayoff
    Nov 22 '18 at 17:51











  • @robmayoff you are right, didn't think it would make a difference

    – Kiril S.
    Nov 23 '18 at 19:25








1




1





You already know about the ?? operator. What is wrong with let str = dict[index] ?? "N/A"?

– rob mayoff
Nov 22 '18 at 17:51





You already know about the ?? operator. What is wrong with let str = dict[index] ?? "N/A"?

– rob mayoff
Nov 22 '18 at 17:51













@robmayoff you are right, didn't think it would make a difference

– Kiril S.
Nov 23 '18 at 19:25





@robmayoff you are right, didn't think it would make a difference

– Kiril S.
Nov 23 '18 at 19:25












3 Answers
3






active

oldest

votes


















4














The issue is that dict[index] is still returning an optional so str has to be an optional. There are two ways you can fix it



You can use the nil coalescing operator to use dict[index] if its not nil otherwise use "N/A"



let str = dict[index] ?? "N/A"


Another option is to use the dictionary default value subscript



let str = dict[index, default: "N/A"]





share|improve this answer
























  • thanks for "default value subscript" option

    – Kiril S.
    Nov 23 '18 at 19:26



















3














You should simply use the nil coalescing operator when defining the str itself. Checking if the value is nil, then still assigning the optional value without unwrapping makes no sense.



let str = dict[index] ?? "N/A"





share|improve this answer
























  • Yes, this works

    – fivewood
    Nov 22 '18 at 18:25



















1














You should use nil coalescing operator - let str = dict[index] ?? "N/A"



Instead of let str = (nil != dict[index]) ? dict[index] : "N/A"



The reason because str is still optional in your case is because dict[index] is optional even if you are checking it to be no nil. So, swift infers the type of str to be String?



You could also force unwrap because you are already checking it to be no nil like -
let str = (nil != dict[index]) ? dict[index]! : "N/A"



But force unwrapping should be avoided whenever possible






share|improve this answer























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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    The issue is that dict[index] is still returning an optional so str has to be an optional. There are two ways you can fix it



    You can use the nil coalescing operator to use dict[index] if its not nil otherwise use "N/A"



    let str = dict[index] ?? "N/A"


    Another option is to use the dictionary default value subscript



    let str = dict[index, default: "N/A"]





    share|improve this answer
























    • thanks for "default value subscript" option

      – Kiril S.
      Nov 23 '18 at 19:26
















    4














    The issue is that dict[index] is still returning an optional so str has to be an optional. There are two ways you can fix it



    You can use the nil coalescing operator to use dict[index] if its not nil otherwise use "N/A"



    let str = dict[index] ?? "N/A"


    Another option is to use the dictionary default value subscript



    let str = dict[index, default: "N/A"]





    share|improve this answer
























    • thanks for "default value subscript" option

      – Kiril S.
      Nov 23 '18 at 19:26














    4












    4








    4







    The issue is that dict[index] is still returning an optional so str has to be an optional. There are two ways you can fix it



    You can use the nil coalescing operator to use dict[index] if its not nil otherwise use "N/A"



    let str = dict[index] ?? "N/A"


    Another option is to use the dictionary default value subscript



    let str = dict[index, default: "N/A"]





    share|improve this answer













    The issue is that dict[index] is still returning an optional so str has to be an optional. There are two ways you can fix it



    You can use the nil coalescing operator to use dict[index] if its not nil otherwise use "N/A"



    let str = dict[index] ?? "N/A"


    Another option is to use the dictionary default value subscript



    let str = dict[index, default: "N/A"]






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 22 '18 at 17:52









    Craig SiemensCraig Siemens

    9,43111941




    9,43111941













    • thanks for "default value subscript" option

      – Kiril S.
      Nov 23 '18 at 19:26



















    • thanks for "default value subscript" option

      – Kiril S.
      Nov 23 '18 at 19:26

















    thanks for "default value subscript" option

    – Kiril S.
    Nov 23 '18 at 19:26





    thanks for "default value subscript" option

    – Kiril S.
    Nov 23 '18 at 19:26













    3














    You should simply use the nil coalescing operator when defining the str itself. Checking if the value is nil, then still assigning the optional value without unwrapping makes no sense.



    let str = dict[index] ?? "N/A"





    share|improve this answer
























    • Yes, this works

      – fivewood
      Nov 22 '18 at 18:25
















    3














    You should simply use the nil coalescing operator when defining the str itself. Checking if the value is nil, then still assigning the optional value without unwrapping makes no sense.



    let str = dict[index] ?? "N/A"





    share|improve this answer
























    • Yes, this works

      – fivewood
      Nov 22 '18 at 18:25














    3












    3








    3







    You should simply use the nil coalescing operator when defining the str itself. Checking if the value is nil, then still assigning the optional value without unwrapping makes no sense.



    let str = dict[index] ?? "N/A"





    share|improve this answer













    You should simply use the nil coalescing operator when defining the str itself. Checking if the value is nil, then still assigning the optional value without unwrapping makes no sense.



    let str = dict[index] ?? "N/A"






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 22 '18 at 17:52









    Dávid PásztorDávid Pásztor

    22.6k83051




    22.6k83051













    • Yes, this works

      – fivewood
      Nov 22 '18 at 18:25



















    • Yes, this works

      – fivewood
      Nov 22 '18 at 18:25

















    Yes, this works

    – fivewood
    Nov 22 '18 at 18:25





    Yes, this works

    – fivewood
    Nov 22 '18 at 18:25











    1














    You should use nil coalescing operator - let str = dict[index] ?? "N/A"



    Instead of let str = (nil != dict[index]) ? dict[index] : "N/A"



    The reason because str is still optional in your case is because dict[index] is optional even if you are checking it to be no nil. So, swift infers the type of str to be String?



    You could also force unwrap because you are already checking it to be no nil like -
    let str = (nil != dict[index]) ? dict[index]! : "N/A"



    But force unwrapping should be avoided whenever possible






    share|improve this answer




























      1














      You should use nil coalescing operator - let str = dict[index] ?? "N/A"



      Instead of let str = (nil != dict[index]) ? dict[index] : "N/A"



      The reason because str is still optional in your case is because dict[index] is optional even if you are checking it to be no nil. So, swift infers the type of str to be String?



      You could also force unwrap because you are already checking it to be no nil like -
      let str = (nil != dict[index]) ? dict[index]! : "N/A"



      But force unwrapping should be avoided whenever possible






      share|improve this answer


























        1












        1








        1







        You should use nil coalescing operator - let str = dict[index] ?? "N/A"



        Instead of let str = (nil != dict[index]) ? dict[index] : "N/A"



        The reason because str is still optional in your case is because dict[index] is optional even if you are checking it to be no nil. So, swift infers the type of str to be String?



        You could also force unwrap because you are already checking it to be no nil like -
        let str = (nil != dict[index]) ? dict[index]! : "N/A"



        But force unwrapping should be avoided whenever possible






        share|improve this answer













        You should use nil coalescing operator - let str = dict[index] ?? "N/A"



        Instead of let str = (nil != dict[index]) ? dict[index] : "N/A"



        The reason because str is still optional in your case is because dict[index] is optional even if you are checking it to be no nil. So, swift infers the type of str to be String?



        You could also force unwrap because you are already checking it to be no nil like -
        let str = (nil != dict[index]) ? dict[index]! : "N/A"



        But force unwrapping should be avoided whenever possible







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 22 '18 at 17:52









        AakashAakash

        1,378715




        1,378715






























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