Converting an Int64
to Int
by passing the Int64
value to the Int
initializer will always work on a 64-bit machine, and it will crash on a 32-bit machine if the integer is outside of the range Int32.min ... Int32.max
.
For safety use the init(truncatingIfNeeded:)
initializer (formerly known as init(truncatingBitPattern:)
in earlier Swift versions) to convert the value:
return Int(truncatingIfNeeded: rowid)
On a 64-bit machine, the truncatingIfNeeded
will do nothing; you will just get an Int
(which is the same size as an Int64
anyway).
On a 32-bit machine, this will throw away the top 32 bits, but it they are all zeroes, then you haven't lost any data. So as long as your value will fit into a 32-bit Int
, you can do this without losing data. If your value is outside of the range Int32.min ... Int32.max
, this will change the value of the Int64
into something that fits in a 32-bit Int
, but it will not crash.
You can see how this works in a Playground. Since Int
in a Playground is a 64-bit Int
, you can explicitly use an Int32
to simulate the behavior of a 32-bit system.
let i: Int64 = 12345678901 // value bigger than maximum 32-bit Int
let j = Int32(truncatingIfNeeded: i) // j = -539,222,987
let k = Int32(i) // crash!
Update for Swift 3/4
In addition to init(truncatingIfNeeded:)
which still works, Swift 3 introduces failable initializers to safely convert one integer type to another. By using init?(exactly:)
you can pass one type to initialize another, and it returns nil
if the initialization fails. The value returned is an optional which must be unwrapped in the usual ways.
For example:
let i: Int64 = 12345678901
if let j = Int32(exactly: i) {
print("\(j) fits into an Int32")
} else {
// the initialization returned nil
print("\(i) is too large for Int32")
}
This allows you to apply the nil coalescing operator to supply a default value if the conversion fails:
// return 0 if rowid is too big to fit into an Int on this device
return Int(exactly: rowid) ?? 0