Cannot divide and assign Int64 value

Question

I was translating this script here to Swift: https://stackoverflow.com/a/3758880/2430555

The error: binary operator '<' cannot be applied to operands of type '()' and 'Int'

But I am stuck in the part below, this is not compiling because \= is not returning the value, instead it's returning (). Any idea what this could be?

var b: Int64 = Int64(1e3)
let bb = ((b/=1000) < 999_950)

Thank you all.

@Alexander-ReinstateMonica https://stackoverflow.com/a/3758880/2430555 — Pedro Paulo Amorim, Dec 09 '19 at 11:23
Ok, good. Here's my take on implementing this, btw https://repl.it/@alexandermomchilov/GhettoByteCountFormatter — Alexander, Dec 09 '19 at 16:25

score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

2

(b/=1000) is a function that returns a void ('aka' () ).

You cant compare a Void and a number (by default)

You can refactor it to:

var b: Int64 = Int64(1e3)
b/=1000
let bb = ( b < 999_950)

Based on the evidence in the other answer by yourself, you are looking for byte converter. You can achieve it like:

func convertBitrateToHumanReadable(bytes: Int64) -> String { ByteCountFormatter().string(fromByteCount: bytes) }

Previously other solution:

You can implement a custom operator for this: (I'm not fan of using this. see @Alexander - Reinstate Monica comment)

infix operator /=> : MultiplicationPrecedence
precedencegroup MultiplicationPrecedence {
  associativity: left
  higherThan: AdditionPrecedence
}

public func /=><T: Numeric>(lhs: inout T, rhs: T) -> T{
    lhs = lhs + rhs
    return lhs
}

Now you can use it like:

let bb = ( b/=>1000 < 999_950)

edited Jun 20 '20 at 09:12

Community

1
1

answered Dec 06 '19 at 18:34

Mojtaba Hosseini

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`b` has to be modified – Pedro Paulo Amorim Dec 06 '19 at 18:36
Updated to what it should be ;) – Mojtaba Hosseini Dec 06 '19 at 18:37
Ok, this is an alternative but it will make my code massive because I will need to wrap everything around functions. Any other idea? – Pedro Paulo Amorim Dec 06 '19 at 18:37
You can have an extension for that. Need a hand for that too? – Mojtaba Hosseini Dec 06 '19 at 18:39
No, I will post my solution down below. – Pedro Paulo Amorim Dec 06 '19 at 18:40
I have added a simple trick for that. Hope it helps you less code. ;) – Mojtaba Hosseini Dec 06 '19 at 18:49
Nice solution! Thanks! – Pedro Paulo Amorim Dec 06 '19 at 18:50
Bleh, don't do this. `/=` (and all other assignment operators) don't return a value for a reason. Not only does it prevent `if x = y` errors in the case of regular assignment, but it also removes all this clever but unreadable stuff here. – Alexander Dec 06 '19 at 18:54

Pedro Paulo Amorim · Answer 2 · 2019-12-06T19:22:02.810

0

Based on @Mojtaba Hosseini answer, I build this (still not working as expected):

public func humanReadableByteCountSI(_ bytes: Int64) -> String {
    let s: String = bytes < 0 ? "-" : ""
    var b: Int64 = bytes == Int64.min ? Int64.max : abs(bytes)
    let d: Int64 = Int64(1e3)

    let div: () -> Int64 = {
        b/=1000
        return b
    }

    return b < 1000 ? String(format: "%i B", bytes)
        : b < 999_950 ? String(format: "%s%.1f kB", s, Float(b / d))
        : div() < 999_950 ? String(format: "%s%.1f MB", s, Float(b / d))
        : div() < 999_950 ? String(format: "%s%.1f GB", s, Float(b / d))
        : div() < 999_950 ? String(format: "%s%.1f TB", s, Float(b / d))
        : div() < 999_950 ? String(format: "%s%.1f PB", s, Float(b / d))
        : String(format: "%s%.1f EB", s, Float(b / Int64(1e6)))
}

edited Dec 06 '19 at 19:22

answered Dec 06 '19 at 18:41

Pedro Paulo Amorim

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Oh I have a very simpler solution for this. [Check this out](https://stackoverflow.com/a/56237217/5623035) – Mojtaba Hosseini Dec 06 '19 at 18:52
1

Before you start implementing something like this, check that something like this exists. In this case, yes, it does: ByteCountFormatter – Alexander Dec 06 '19 at 18:53
`var b: Int64 = bytes == Int64.min ? Int64.max : abs(bytes)` what ಠ_ಠ – Alexander Dec 06 '19 at 18:56
You also have a critical memory error in `String(format: "%s%.1f B", bytes)`, because you forgot the `s` parameter. The `bytes` is read as a string (undefined behaviour), followed by whatever else in memory, read as a float (also undefined behaviour). Such is the risk of using legacy C/ObjC APIs. – Alexander Dec 06 '19 at 18:58
@MojtabaHosseini I simply implemented this for the challenge of converting that Java code to Swift. – Pedro Paulo Amorim Dec 06 '19 at 19:19
@Alexander-ReinstateMonica I Hope I got it sorted. – Pedro Paulo Amorim Dec 06 '19 at 19:24

Cannot divide and assign Int64 value

2 Answers2

Based on the evidence in the other answer by yourself, you are looking for byte converter. You can achieve it like:

Previously other solution:

Linked