Amit Nivedan Kalra - 1 month ago 26

Swift Question

I'm trying to convert Binary and stuff with Swift and this is my code :

`let hexa = String(Int(a, radix: 2)!, radix: 16)// Converting binary to hexadecimal`

I am getting the error

Cannot convert value type of 'Int' to expected argument type 'String'

Answer

You're misunderstanding how integers are stored.

There is no notion of a "decimal" `Int`

, a "hexadecimal" `Int`

, etc. When you have an `Int`

in memory, it's always binary (radix 2). It's stored as a series of 64 or 32 bits.

When you try to assign to the `Int`

a value like `10`

(decimal), `0xA`

(hex), `0b1010`

(binary), the compiler does the necessary parsing to convert your source code's string representation of that `Int`

, into a series of bits that can be stored in the `Int`

's 64 or 32 bits of memory.

When you try to use the `Int`

, for example with `print(a)`

, there is conversion behind the scenes to take that `Int`

's binary representation in memory, and convert it into a `String`

whose symbols represent an `Int`

in base 10, using the symbols we're used to (`0-9`

).

On a more fundamental, it helps to understand that the notion of a radix is a construct devised purely for our convenience when working with numbers. Abstractly, a number has a magnitude that is a distinct entity, uncoupled from any radix. A magnitude can be represented concretely using a textual representation and a radix.

This part `Int(a, radix: 2)`

, doesn't make sense. Even supposing such an initializer (`Int.init?(Int, radix: Int)`

) existed, it wouldn't do anything!. If `a = 5`

, then `a`

is stored as binary `0b101`

. This would then be parsed from binary into an Int, giving you... `0b101`

, or the same `5`

you started with.

On the other hand, `Strings`

can have a notion of a radix, because they can be a textual representation of a decimal `Int`

, a hex `Int`

, etc. To convert from a `String`

that contains a number, you use `Int.init?(String, radix: Int)`

. The key here is that it takes a ** String** parameter.

```
let a = 10 //decimal 10 is stored as binary in memory 1010
let hexa = String(a, radix: 16) //the Int is converted to a string, 0xA
```

Source (Stackoverflow)

Comments