'Cannot convert value of type Substring to expected argument type String - Swift 4
Solution 1:[1]
my two cents for serro in different context. I was trying to get an array of "String" splitting a string. "split" gives back "Substring", for efficiency reason (as per Swift.org litre).
So I ended up doing:
let buffer = "one,two,three"
let rows = buffer.split(separator:",")
let realStrings = rows.map { subString -> String in
return String(subString)
}
print(realStrings)
Ape can help someone else.
Sources
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Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | ingconti |

