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Louis Guillard
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Asked: 2022-05-12T09:14:31+00:00 In:

Fixing the error: swift type of expression is ambiguous without more context

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Hi developer experts, I have a small but frustrating use case, and so far, I couldn’t get my head around this problem & ideal solution. I am running my command, and I am facing one problem with the swift type of expression is ambiguous without more context Below is the command I used:

let imageToDeleteParameters = imagesToDelete.map { ["id": $0.id, "url": $0.url.absoluteString, "_destroy": true] }
extension TutorialCreationRequest: WebserviceParametrable {
  func toParameters() -> [String: AnyObject] {
  let imageParameters = images.map { ["url": $0] }
  let imageToDeleteParameters = imagesToDelete.map { ["id": $0.id, "url": $0.url.absoluteString, "_destroy": true] }
  return [
  "title": title,
  "is_draft": isDraft,
  "difficulty": difficulty,
  "duration": duration,
  "cost": cost,
  "user_id": userId,
  "description": description,
  "to_sell": toSell,
  "images": [imageParameters, imageToDeleteParameters].flatMap { $0 }
  ]
  }
 }

When I run it, I get the following error:

Type of expression is ambiguous without more context

I am looking forward to gaining some knowledge from all experts. Thank you, guys!

swift type of expression is ambiguous
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2 Answers

  1. Best Answer
    Expert

    The cause: The reason is that you you changed the name of a Variable and did not refactor when doing that. Solution: It will work if you declare clearly the inputs and then mapping task do the following way: 

    Replace the real class of “$0” for “WhateverClass” in that code snippet. Fixed code: 

    let imageToDeleteParameters = imagesToDelete.map { (whatever : WhateverClass) -> Dictionary<String, Any> in ["id": whatever.id, "url": whatever.url.absoluteString, "_destroy": true] }
    • 12

  2. This is when you have an argument name that is not correct for a function.

    Example: 

    functionWithArguments(argumentNameWrong: , argumentName2: )

    And You declared your function as 

    functionWithArguments(argumentName1: , argumentName2: ){}

    This happens most often when you change the name of a Variable. When you do this, make sure to refactor.

    • 19

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