Type mismatch: inferred type is Double but (Double) -> Double was expected-CodePudding

I am working on a solution for the Temperature Control Problem. It is one of the problems in Kotlin learning fundamentals. So for the thrid argument inside printFinalTemprature function, I am passing a function called conversionFormula but what ever I do, I kept getting the type mismatch error. What am I missing?

fun main() {
    var temp : Double = 10.0
    fun celsiusToFahrenheit(celsius : Double): Double {
        return 9/5 * (celsius)   32
    }
    fun kelvinToCelsius(kelvin  : Double):Double {
        return kelvin - 273.15
    } 
    fun conversionFormula(fahrenheit: Double): Double {
        var fiveNine :Double = 0.56 
        var result :Double = (fiveNine*(fahrenheit - 32.0)   273.15)
        return result
    } 
   printFinalTemperature(temp, "Kelvin", "Celsius", conversionFormula(temp))
}


fun printFinalTemperature(
    initialMeasurement: Double, 
    initialUnit: String, 
    finalUnit: String, 
    conversionFormula: (Double) -> Double
) {
    val finalMeasurement = String.format("%.2f", conversionFormula(initialMeasurement)) // two decimal places
    println("$initialMeasurement degrees $initialUnit is $finalMeasurement degrees $finalUnit.")
}

CodePudding user response：

You just need to switch how you are passing the function. Everything is an expression in Kotlin, and since your printFinalTemperature expects a function as the final argument, you have 2 ways to accomplish this.

The first is to use the function reference with ::. This will only work if the function you are referencing has the same expected signature, so in this case (Double) -> Double. You can try changing one of the Double to Int and you'll see how it breaks.

printFinalTemperature(temp, "Kelvin", "Celsius", ::conversionFormula)

Secondarily, you can pass in a lambda as the final argument, allowing it to "infer" that the function is the correct signature.

printFinalTemperature(temp, "Kelvin", "Celsius") {
    conversionFormula(it)
}

In this case, the it comes from the lambda as the default argument name. You could give it an explicit name such as:

printFinalTemperature(temp, "Kelvin", "Celsius") { newTemp ->
    conversionFormula(newtemp)
}