C# in Depth

250 CHAPTER 9 Lambda expressions and expression trees1 Phase 1 begins.2 The first parameter is of type TInput, and the first argument is of type string.We infer that there must be an implicit conversion from string to TInput.3 The second parameter is of type Converter, and the secondargument is an implicitly typed lambda expression. No inference is performed—wedon’t have enough information.4 Phase 2 begins.5 TInput doesn’t depend on any unfixed type parameters, so it’s fixed to string.6 The second argument now has a fixed input type, but an unfixed output type. Wecan consider it to be (string x) => x.Length and infer the return type as int.Therefore an implicit conversion must take place from int to TOutput.7 Phase 2 repeats.8 TOutput doesn’t depend on anything unfixed, so it’s fixed to int.9 There are now no unfixed type parameters, so inference succeeds.Complicated, eh? Still, it does the job—the result is what we’d want (TInput=string,TOutput=int) and everything compiles without any problems. The importance ofphase 2 repeating is best shown with another example, however. Listing 9.15 shows twoconversions being performed, with the output of the first one becoming the input ofthe second. Until we’ve worked out the output type of the first conversion, we don’tknow the input type of the second, so we can’t infer its output type either.Listing 9.15Multistage type inferencestatic void ConvertTwice(TInput input,Converter firstConversion,Converter secondConversion){TMiddle middle = firstConversion(input);TOutput output = secondConversion(middle);Console.WriteLine(output);}...ConvertTwice("Another string",text => text.Length,length => Math.Sqrt(length));The first thing to notice is that the method signature appears to be pretty horrific. It’snot too bad when you stop being scared and just look at it carefully—and certainly theexample usage makes it more obvious. We take a string, and perform a conversion onit: the same conversion as before, just a length calculation. We then take that length(an int) and find its square root (a double).Phase 1 of type inference tells the compiler that there must be a conversion fromstring to TInput. The first time through phase 2, TInput is fixed to string and weinfer that there must be a conversion from int to TMiddle. The second time throughLicensed to Rhona Hadida