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I know in Swift one way of generating a random number is:

let getXOrY = random(2)

However I am getting the error:

Type '()' does not conform to protocol 'IntegerLiteralConvertible'

I have also tried:

let getXOrY = arc4random(2)

and it creates the same error.

I am not sure why this is happening. In other responses on this site it seemed to be as easy as that one line of code. Also, please explain the error means. Thank you.

Acoop
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1 Answers1

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This is because the random number generator is arc4random_uniform() or arc4random().

More on this answer here: How does one generate a random number in Apple's Swift language?

Community
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nonamorando
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