Okay, I am getting this strange error in xcode 8. In the below code, which is for dialing a number,:
@IBAction func dialNumber(_ sender: AnyObject) {
if let url = URL(string: "tel:/"tel://\(8708382937)") {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
}
}
I am getting the error: "Integer literal '8708382937' overflows when stored into 'Int"' On line 02.
This code was working just fine in xcode 7. Any help?
It happens when your target is a 32-bit device (such as iPhone 5). There are several workarounds:
1. Use a string, since you are hard-coding the value anyway:
if let url = URL(string: "tel://8708382937") {2. Cast it to Int64:
if let url = URL(string: "tel://\(8708382937 as Int64)") {3. Use a variable and specify the type:
let tel: Int64 = 8708382937
if let url = URL(string: "tel://\(tel)") {Swift is inferring a type of Int for the integer literal, which has a max value of 2147483647 on a 32-bit system.
It happens when your target is a 32-bit device (such as iPhone 5). There are several workarounds:
1. Use a string, since you are hard-coding the value anyway:
if let url = URL(string: "tel://8708382937") {2. Cast it to Int64:
if let url = URL(string: "tel://\(8708382937 as Int64)") {3. Use a variable and specify the type:
let tel: Int64 = 8708382937
if let url = URL(string: "tel://\(tel)") {Swift is inferring a type of Int for the integer literal, which has a max value of 2147483647 on a 32-bit system.