Taeho  Sung Taeho Sung - 1 year ago 65
Swift Question

Swift Types: combining integer and floating-point literals in an expression

I'm new to studying swift and iOS and read this in their documentation:

If you combine integer and floating-point literals in an expression, a type of Double will be inferred from the context:

let anotherPi = 3 + 0.14159
// anotherPi is also inferred to be of type Double

The literal value of 3 has no explicit type in and of itself, and so
an appropriate output type of Double is inferred from the presence of
a floating-point literal as part of the addition.

I understand that if you try adding an explicit integer type and an explicit double type, swift will give you an error because you can't add an int and a double in swift. Does the above quote from the documentation mean that the integer literal 3 is inferred to be of type double by swift and then added to the floating literal 0.14159 which is also inferred to be of type double? or does it just mean that if you add an int literal and a floating point literal the output is inferred to be a double?

Answer Source

Floating-point literals are inferred to be Double by default. You can test this in a Playground:

// Swift 3

let a = 0.14159

print(type(of: a)) // Double

let b = 1 + a // works

let c = Int(1) + a // doesn't work

Example B works because the integer literal 1 can be inferred to be Double (since it is being added to a Double). Example C doesn't work because 1 is an explicit Int. This preserves Swift's numerical type safety while still making it easy to perform simple math.

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