## SOLVED: Does reduce() use short-circuit evaluation?

 Mar '22 As I understand things, the compiler will short-circuit evaluations once it nows the outcome. For example `if (a && b && c && d)` if a is false, there is no need to evaluate b, c or d as regardless of their value, the if will always be false. Now, assume there is an array of Booleans and you want to find out if they are all true. You could do that with a reduce() statement. `array.reduce(true){\$0 && \$1}` Does the evaluation of the reduce stop once a false value is encountered? FWIW, I came up with this while writing a function to convert Apples OSErr type to a printable string if possible. ``````// convert Apple 32-bit int to 4 characters if valid // 1819304813 = "lpcm" // 1718449215 = "fmt?" // -50 = (-50) func int32ToString(_ error: OSStatus) -> String { let chars = Swift.withUnsafeBytes(of: error.bigEndian){ Data(\$0) }.map{ Character(Unicode.Scalar(\$0)) } return chars.reduce(true){ \$0 && \$1.isASCII } ? "\"\(String(chars))\"" : "(\(error))" }`````` I like eliminating For loops with map() or reduce() statements. 2 Mar '22 `.reduce` works on the sequence in the array, and processes all the elements. If you try this in Playground, you can see that `reduce` is acted upon 11 times. ``````let boolArray = [false, true, true, true, false, true, false, false, true, true] let reducedBoolArray = boolArray.reduce(true) {\$0 && \$1} print ("Bool Array", boolArray) print ("Reduced to", reducedBoolArray)`````` 3 Mar '22 Does the evaluation of the reduce stop once a false value is encountered? The `reduce` isn't evaluating a condition, it's just assembling a value based on what you give it. It will go through all the values in the source until it's finished and then present to you the final value. So, no. 2

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