Argument labels '(stringInterpolationSegment:)' do not match any available overloadsGMStepper - Argument labels '(stringInterpolationSegment:)' do not match any available overloads@selector() in Swift?swift 3 error : Argument labels '(_:)' do not match any available overloadsSwift CGPoint Argument labels '(_:, _:)' do not match any available overloadsNumberFormatter Argument labels '(_:)' do not match any available overloadsArgument labels '(_:)' do not match any available overloadsArgument labels '(UUIDString)' do not match any available overloadsArgument labels '(_:, _:)' do not match any available overloadsArgument Labels '(collectionviewlayout:)' do not match any available overloadsError: Argument labels '(_:, _:)' do not match any available overloadsswift cgsize error Argument labels '(_:, _:)' do not match any available overloads

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Argument labels '(stringInterpolationSegment:)' do not match any available overloads


GMStepper - Argument labels '(stringInterpolationSegment:)' do not match any available overloads@selector() in Swift?swift 3 error : Argument labels '(_:)' do not match any available overloadsSwift CGPoint Argument labels '(_:, _:)' do not match any available overloadsNumberFormatter Argument labels '(_:)' do not match any available overloadsArgument labels '(_:)' do not match any available overloadsArgument labels '(UUIDString)' do not match any available overloadsArgument labels '(_:, _:)' do not match any available overloadsArgument Labels '(collectionviewlayout:)' do not match any available overloadsError: Argument labels '(_:, _:)' do not match any available overloadsswift cgsize error Argument labels '(_:, _:)' do not match any available overloads






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1















After updating the latest Xcode Version 10.2 (10E125), I am getting this error :



Argument labels '(stringInterpolationSegment:)' do not match any available overloads



Could not find any solution yet, any idea please ?



let interpolation1 = String(stringInterpolationSegment:self.addSpotAnnotation!.coordinate.latitude)
let interpolation2 = String(stringInterpolationSegment:self.addSpotAnnotation!.coordinate.longitude)
let coordinate:String = interpolation1 + "," + interpolation2









share|improve this question


























  • Can you provide the line of code, or lines, where this error is occurring?

    – Mike Taverne
    Mar 27 at 4:38












  • Its an existing app and there are lots of place where its been used. One of them is like this: let coordinate:String = String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude) + "," + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)

    – Arafin Russell
    Mar 27 at 4:50







  • 1





    Replace stringInterpolationSegment with stringInterpolation.

    – Sachin Vas
    Mar 27 at 5:17

















1















After updating the latest Xcode Version 10.2 (10E125), I am getting this error :



Argument labels '(stringInterpolationSegment:)' do not match any available overloads



Could not find any solution yet, any idea please ?



let interpolation1 = String(stringInterpolationSegment:self.addSpotAnnotation!.coordinate.latitude)
let interpolation2 = String(stringInterpolationSegment:self.addSpotAnnotation!.coordinate.longitude)
let coordinate:String = interpolation1 + "," + interpolation2









share|improve this question


























  • Can you provide the line of code, or lines, where this error is occurring?

    – Mike Taverne
    Mar 27 at 4:38












  • Its an existing app and there are lots of place where its been used. One of them is like this: let coordinate:String = String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude) + "," + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)

    – Arafin Russell
    Mar 27 at 4:50







  • 1





    Replace stringInterpolationSegment with stringInterpolation.

    – Sachin Vas
    Mar 27 at 5:17













1












1








1








After updating the latest Xcode Version 10.2 (10E125), I am getting this error :



Argument labels '(stringInterpolationSegment:)' do not match any available overloads



Could not find any solution yet, any idea please ?



let interpolation1 = String(stringInterpolationSegment:self.addSpotAnnotation!.coordinate.latitude)
let interpolation2 = String(stringInterpolationSegment:self.addSpotAnnotation!.coordinate.longitude)
let coordinate:String = interpolation1 + "," + interpolation2









share|improve this question
















After updating the latest Xcode Version 10.2 (10E125), I am getting this error :



Argument labels '(stringInterpolationSegment:)' do not match any available overloads



Could not find any solution yet, any idea please ?



let interpolation1 = String(stringInterpolationSegment:self.addSpotAnnotation!.coordinate.latitude)
let interpolation2 = String(stringInterpolationSegment:self.addSpotAnnotation!.coordinate.longitude)
let coordinate:String = interpolation1 + "," + interpolation2






swift xcode10.2






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 27 at 9:10









Sachin Vas

1,5468 silver badges15 bronze badges




1,5468 silver badges15 bronze badges










asked Mar 27 at 3:39









Arafin RussellArafin Russell

5477 silver badges21 bronze badges




5477 silver badges21 bronze badges















  • Can you provide the line of code, or lines, where this error is occurring?

    – Mike Taverne
    Mar 27 at 4:38












  • Its an existing app and there are lots of place where its been used. One of them is like this: let coordinate:String = String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude) + "," + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)

    – Arafin Russell
    Mar 27 at 4:50







  • 1





    Replace stringInterpolationSegment with stringInterpolation.

    – Sachin Vas
    Mar 27 at 5:17

















  • Can you provide the line of code, or lines, where this error is occurring?

    – Mike Taverne
    Mar 27 at 4:38












  • Its an existing app and there are lots of place where its been used. One of them is like this: let coordinate:String = String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude) + "," + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)

    – Arafin Russell
    Mar 27 at 4:50







  • 1





    Replace stringInterpolationSegment with stringInterpolation.

    – Sachin Vas
    Mar 27 at 5:17
















Can you provide the line of code, or lines, where this error is occurring?

– Mike Taverne
Mar 27 at 4:38






Can you provide the line of code, or lines, where this error is occurring?

– Mike Taverne
Mar 27 at 4:38














Its an existing app and there are lots of place where its been used. One of them is like this: let coordinate:String = String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude) + "," + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)

– Arafin Russell
Mar 27 at 4:50






Its an existing app and there are lots of place where its been used. One of them is like this: let coordinate:String = String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude) + "," + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)

– Arafin Russell
Mar 27 at 4:50





1




1





Replace stringInterpolationSegment with stringInterpolation.

– Sachin Vas
Mar 27 at 5:17





Replace stringInterpolationSegment with stringInterpolation.

– Sachin Vas
Mar 27 at 5:17












3 Answers
3






active

oldest

votes


















2














The error is due to changes to how string interpolation works in Swift 5.



The solution is not to replace String(stringInterpolationSegment:) with String(stringInterpolation:):




We do not propose preserving existing init(stringInterpolation:) or init(stringInterpolationSegment:) initializers, since they have always been documented as calls that should not be used directly. [emphasis added]




The example you gave:



coordinate:String = 
String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude)
+ ","
+ String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)


can much more easily be written as:



let coordinate = "(self.addSpotAnnotation!.coordinate.latitude),(self.addSpotAnnotation!.coordinate.longitude)"





share|improve this answer
































    1














    String Interpolation Updates



    Swift 4.2 implements string interpolation by interpolating segments:



    let language = "Swift"
    let languageSegment = String(stringInterpolationSegment: language)
    let space = " "
    let spaceSegment = String(stringInterpolationSegment: space)
    let version = 4.2
    let versionSegment = String(stringInterpolationSegment: version)
    let string = String(stringInterpolation: languageSegment, spaceSegment, versionSegment)


    In this code, the compiler first wraps each literal segment and then interpolates one with init(stringInterpolationSegment:) . Then, it wraps all segments together with init(stringInterpolation:)



    Swift 5 takes a completely different approach



    // 1
    var interpolation = DefaultStringInterpolation(
    literalCapacity: 7,
    interpolationCount: 1)
    // 2
    let language = "Swift"
    interpolation.appendLiteral(language)
    let space = " "
    interpolation.appendLiteral(space)
    let version = 5
    interpolation.appendInterpolation(version)
    // 3
    let string = String(stringInterpolation: interpolation)


    Here’s what this code does:



    Define a DefaultStringInterpolation instance with a certain capacity and interpolation count.
    Call appendLiteral(:) or appendInterpolation(:) to add literals and interpolated values to interpolation.
    Produce the final interpolated string by calling init(stringInterpolation:)



    credit: raywenderlich






    share|improve this answer


































      1














      public func == <T>(lhs: ResultTest<T>, rhs: ResultTest<T>) -> Bool 
      return "(lhs)" == "(rhs)"






      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









        2














        The error is due to changes to how string interpolation works in Swift 5.



        The solution is not to replace String(stringInterpolationSegment:) with String(stringInterpolation:):




        We do not propose preserving existing init(stringInterpolation:) or init(stringInterpolationSegment:) initializers, since they have always been documented as calls that should not be used directly. [emphasis added]




        The example you gave:



        coordinate:String = 
        String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude)
        + ","
        + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)


        can much more easily be written as:



        let coordinate = "(self.addSpotAnnotation!.coordinate.latitude),(self.addSpotAnnotation!.coordinate.longitude)"





        share|improve this answer





























          2














          The error is due to changes to how string interpolation works in Swift 5.



          The solution is not to replace String(stringInterpolationSegment:) with String(stringInterpolation:):




          We do not propose preserving existing init(stringInterpolation:) or init(stringInterpolationSegment:) initializers, since they have always been documented as calls that should not be used directly. [emphasis added]




          The example you gave:



          coordinate:String = 
          String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude)
          + ","
          + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)


          can much more easily be written as:



          let coordinate = "(self.addSpotAnnotation!.coordinate.latitude),(self.addSpotAnnotation!.coordinate.longitude)"





          share|improve this answer



























            2












            2








            2







            The error is due to changes to how string interpolation works in Swift 5.



            The solution is not to replace String(stringInterpolationSegment:) with String(stringInterpolation:):




            We do not propose preserving existing init(stringInterpolation:) or init(stringInterpolationSegment:) initializers, since they have always been documented as calls that should not be used directly. [emphasis added]




            The example you gave:



            coordinate:String = 
            String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude)
            + ","
            + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)


            can much more easily be written as:



            let coordinate = "(self.addSpotAnnotation!.coordinate.latitude),(self.addSpotAnnotation!.coordinate.longitude)"





            share|improve this answer













            The error is due to changes to how string interpolation works in Swift 5.



            The solution is not to replace String(stringInterpolationSegment:) with String(stringInterpolation:):




            We do not propose preserving existing init(stringInterpolation:) or init(stringInterpolationSegment:) initializers, since they have always been documented as calls that should not be used directly. [emphasis added]




            The example you gave:



            coordinate:String = 
            String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.latitude)
            + ","
            + String(stringInterpolationSegment: self.addSpotAnnotation!.coordinate.longitude)


            can much more easily be written as:



            let coordinate = "(self.addSpotAnnotation!.coordinate.latitude),(self.addSpotAnnotation!.coordinate.longitude)"






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Mar 27 at 5:43









            Mike TaverneMike Taverne

            6,6482 gold badges24 silver badges42 bronze badges




            6,6482 gold badges24 silver badges42 bronze badges


























                1














                String Interpolation Updates



                Swift 4.2 implements string interpolation by interpolating segments:



                let language = "Swift"
                let languageSegment = String(stringInterpolationSegment: language)
                let space = " "
                let spaceSegment = String(stringInterpolationSegment: space)
                let version = 4.2
                let versionSegment = String(stringInterpolationSegment: version)
                let string = String(stringInterpolation: languageSegment, spaceSegment, versionSegment)


                In this code, the compiler first wraps each literal segment and then interpolates one with init(stringInterpolationSegment:) . Then, it wraps all segments together with init(stringInterpolation:)



                Swift 5 takes a completely different approach



                // 1
                var interpolation = DefaultStringInterpolation(
                literalCapacity: 7,
                interpolationCount: 1)
                // 2
                let language = "Swift"
                interpolation.appendLiteral(language)
                let space = " "
                interpolation.appendLiteral(space)
                let version = 5
                interpolation.appendInterpolation(version)
                // 3
                let string = String(stringInterpolation: interpolation)


                Here’s what this code does:



                Define a DefaultStringInterpolation instance with a certain capacity and interpolation count.
                Call appendLiteral(:) or appendInterpolation(:) to add literals and interpolated values to interpolation.
                Produce the final interpolated string by calling init(stringInterpolation:)



                credit: raywenderlich






                share|improve this answer































                  1














                  String Interpolation Updates



                  Swift 4.2 implements string interpolation by interpolating segments:



                  let language = "Swift"
                  let languageSegment = String(stringInterpolationSegment: language)
                  let space = " "
                  let spaceSegment = String(stringInterpolationSegment: space)
                  let version = 4.2
                  let versionSegment = String(stringInterpolationSegment: version)
                  let string = String(stringInterpolation: languageSegment, spaceSegment, versionSegment)


                  In this code, the compiler first wraps each literal segment and then interpolates one with init(stringInterpolationSegment:) . Then, it wraps all segments together with init(stringInterpolation:)



                  Swift 5 takes a completely different approach



                  // 1
                  var interpolation = DefaultStringInterpolation(
                  literalCapacity: 7,
                  interpolationCount: 1)
                  // 2
                  let language = "Swift"
                  interpolation.appendLiteral(language)
                  let space = " "
                  interpolation.appendLiteral(space)
                  let version = 5
                  interpolation.appendInterpolation(version)
                  // 3
                  let string = String(stringInterpolation: interpolation)


                  Here’s what this code does:



                  Define a DefaultStringInterpolation instance with a certain capacity and interpolation count.
                  Call appendLiteral(:) or appendInterpolation(:) to add literals and interpolated values to interpolation.
                  Produce the final interpolated string by calling init(stringInterpolation:)



                  credit: raywenderlich






                  share|improve this answer





























                    1












                    1








                    1







                    String Interpolation Updates



                    Swift 4.2 implements string interpolation by interpolating segments:



                    let language = "Swift"
                    let languageSegment = String(stringInterpolationSegment: language)
                    let space = " "
                    let spaceSegment = String(stringInterpolationSegment: space)
                    let version = 4.2
                    let versionSegment = String(stringInterpolationSegment: version)
                    let string = String(stringInterpolation: languageSegment, spaceSegment, versionSegment)


                    In this code, the compiler first wraps each literal segment and then interpolates one with init(stringInterpolationSegment:) . Then, it wraps all segments together with init(stringInterpolation:)



                    Swift 5 takes a completely different approach



                    // 1
                    var interpolation = DefaultStringInterpolation(
                    literalCapacity: 7,
                    interpolationCount: 1)
                    // 2
                    let language = "Swift"
                    interpolation.appendLiteral(language)
                    let space = " "
                    interpolation.appendLiteral(space)
                    let version = 5
                    interpolation.appendInterpolation(version)
                    // 3
                    let string = String(stringInterpolation: interpolation)


                    Here’s what this code does:



                    Define a DefaultStringInterpolation instance with a certain capacity and interpolation count.
                    Call appendLiteral(:) or appendInterpolation(:) to add literals and interpolated values to interpolation.
                    Produce the final interpolated string by calling init(stringInterpolation:)



                    credit: raywenderlich






                    share|improve this answer















                    String Interpolation Updates



                    Swift 4.2 implements string interpolation by interpolating segments:



                    let language = "Swift"
                    let languageSegment = String(stringInterpolationSegment: language)
                    let space = " "
                    let spaceSegment = String(stringInterpolationSegment: space)
                    let version = 4.2
                    let versionSegment = String(stringInterpolationSegment: version)
                    let string = String(stringInterpolation: languageSegment, spaceSegment, versionSegment)


                    In this code, the compiler first wraps each literal segment and then interpolates one with init(stringInterpolationSegment:) . Then, it wraps all segments together with init(stringInterpolation:)



                    Swift 5 takes a completely different approach



                    // 1
                    var interpolation = DefaultStringInterpolation(
                    literalCapacity: 7,
                    interpolationCount: 1)
                    // 2
                    let language = "Swift"
                    interpolation.appendLiteral(language)
                    let space = " "
                    interpolation.appendLiteral(space)
                    let version = 5
                    interpolation.appendInterpolation(version)
                    // 3
                    let string = String(stringInterpolation: interpolation)


                    Here’s what this code does:



                    Define a DefaultStringInterpolation instance with a certain capacity and interpolation count.
                    Call appendLiteral(:) or appendInterpolation(:) to add literals and interpolated values to interpolation.
                    Produce the final interpolated string by calling init(stringInterpolation:)



                    credit: raywenderlich







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Mar 27 at 5:50

























                    answered Mar 27 at 5:39









                    Arafin RussellArafin Russell

                    5477 silver badges21 bronze badges




                    5477 silver badges21 bronze badges
























                        1














                        public func == <T>(lhs: ResultTest<T>, rhs: ResultTest<T>) -> Bool 
                        return "(lhs)" == "(rhs)"






                        share|improve this answer































                          1














                          public func == <T>(lhs: ResultTest<T>, rhs: ResultTest<T>) -> Bool 
                          return "(lhs)" == "(rhs)"






                          share|improve this answer





























                            1












                            1








                            1







                            public func == <T>(lhs: ResultTest<T>, rhs: ResultTest<T>) -> Bool 
                            return "(lhs)" == "(rhs)"






                            share|improve this answer















                            public func == <T>(lhs: ResultTest<T>, rhs: ResultTest<T>) -> Bool 
                            return "(lhs)" == "(rhs)"







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Jun 13 at 7:15

























                            answered Jun 13 at 5:37









                            GiangGiang

                            2,03417 silver badges20 bronze badges




                            2,03417 silver badges20 bronze badges






























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