Swift Hex to byte (iOS)
0
I have commands in hex format (for example 0x01, 0x02, etc)
How to enter them in xcode (what type of variables)? And how to convert them into bytes to calculate the checksum (CRC8)?
ios swift xcode hex crc
asked Nov 16 '18 at 10:53
Dmitriy ToropkinDmitriy Toropkin
31
add a comment |
0
I have commands in hex format (for example 0x01, 0x02, etc)
How to enter them in xcode (what type of variables)? And how to convert them into bytes to calculate the checksum (CRC8)?
ios swift xcode hex crc
asked Nov 16 '18 at 10:53
Dmitriy ToropkinDmitriy Toropkin
31
add a comment |
0
0
0
I have commands in hex format (for example 0x01, 0x02, etc)
How to enter them in xcode (what type of variables)? And how to convert them into bytes to calculate the checksum (CRC8)?
ios swift xcode hex crc
asked Nov 16 '18 at 10:53
Dmitriy ToropkinDmitriy Toropkin
31
I have commands in hex format (for example 0x01, 0x02, etc)
How to enter them in xcode (what type of variables)? And how to convert them into bytes to calculate the checksum (CRC8)?
ios swift xcode hex crc
ios swift xcode hex crc
asked Nov 16 '18 at 10:53
Dmitriy ToropkinDmitriy Toropkin
31
asked Nov 16 '18 at 10:53
Dmitriy ToropkinDmitriy Toropkin
31
asked Nov 16 '18 at 10:53
Dmitriy ToropkinDmitriy Toropkin
31
asked Nov 16 '18 at 10:53
Dmitriy ToropkinDmitriy Toropkin
31
asked Nov 16 '18 at 10:53
Dmitriy ToropkinDmitriy Toropkin
31
31
add a comment |
add a comment |
1 Answer
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You can convert your hexa string back to array of UInt8 using this extension :
extension StringProtocol {
var hexa2Bytes: [UInt8] {
let hexa = Array(self)
return stride(from: 0, to: count, by: 2).compactMap { UInt8(String(hexa[$0..<$0.advanced(by: 2)]), radix: 16) }
}}
answered Nov 16 '18 at 12:32
YD-YD-
19618
you forgot } in the end Thank you, it works great!
– Dmitriy Toropkin
Nov 16 '18 at 12:56
@DmitriyToropkin Good.Mark it as correct answer so others can make use of it.
– YD-
Nov 16 '18 at 13:23
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
You can convert your hexa string back to array of UInt8 using this extension :
extension StringProtocol {
var hexa2Bytes: [UInt8] {
let hexa = Array(self)
return stride(from: 0, to: count, by: 2).compactMap { UInt8(String(hexa[$0..<$0.advanced(by: 2)]), radix: 16) }
}}
answered Nov 16 '18 at 12:32
YD-YD-
19618
you forgot } in the end Thank you, it works great!
– Dmitriy Toropkin
Nov 16 '18 at 12:56
@DmitriyToropkin Good.Mark it as correct answer so others can make use of it.
– YD-
Nov 16 '18 at 13:23
add a comment |
0
You can convert your hexa string back to array of UInt8 using this extension :
extension StringProtocol {
var hexa2Bytes: [UInt8] {
let hexa = Array(self)
return stride(from: 0, to: count, by: 2).compactMap { UInt8(String(hexa[$0..<$0.advanced(by: 2)]), radix: 16) }
}}
answered Nov 16 '18 at 12:32
YD-YD-
19618
you forgot } in the end Thank you, it works great!
– Dmitriy Toropkin
Nov 16 '18 at 12:56
@DmitriyToropkin Good.Mark it as correct answer so others can make use of it.
– YD-
Nov 16 '18 at 13:23
add a comment |
0
0
0
You can convert your hexa string back to array of UInt8 using this extension :
extension StringProtocol {
var hexa2Bytes: [UInt8] {
let hexa = Array(self)
return stride(from: 0, to: count, by: 2).compactMap { UInt8(String(hexa[$0..<$0.advanced(by: 2)]), radix: 16) }
}}
answered Nov 16 '18 at 12:32
YD-YD-
19618
You can convert your hexa string back to array of UInt8 using this extension :
extension StringProtocol {
var hexa2Bytes: [UInt8] {
let hexa = Array(self)
return stride(from: 0, to: count, by: 2).compactMap { UInt8(String(hexa[$0..<$0.advanced(by: 2)]), radix: 16) }
}}
answered Nov 16 '18 at 12:32
YD-YD-
19618
edited Nov 16 '18 at 13:24
answered Nov 16 '18 at 12:32
YD-YD-
19618
answered Nov 16 '18 at 12:32
YD-YD-
19618
answered Nov 16 '18 at 12:32
YD-YD-
19618
19618
you forgot } in the end Thank you, it works great!
– Dmitriy Toropkin
Nov 16 '18 at 12:56
@DmitriyToropkin Good.Mark it as correct answer so others can make use of it.
– YD-
Nov 16 '18 at 13:23
add a comment |
you forgot } in the end Thank you, it works great!
– Dmitriy Toropkin
Nov 16 '18 at 12:56
@DmitriyToropkin Good.Mark it as correct answer so others can make use of it.
– YD-
Nov 16 '18 at 13:23
you forgot } in the end Thank you, it works great!
– Dmitriy Toropkin
Nov 16 '18 at 12:56
you forgot } in the end Thank you, it works great!
– Dmitriy Toropkin
Nov 16 '18 at 12:56
@DmitriyToropkin Good.Mark it as correct answer so others can make use of it.
– YD-
Nov 16 '18 at 13:23
@DmitriyToropkin Good.Mark it as correct answer so others can make use of it.
– YD-
Nov 16 '18 at 13:23
add a comment |
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