How to get a specific character from index of a string in swift

Question

I am trying to build a Binary to Decimal calculator for the Apple Watch using Swift 4.

The code I am having trouble is this:

var i = 0
var labelInputInt = 0
let labelOutputString = "10010" // Random number in binary 
let reverse = String(labelOutputString.reversed()) // Reversing the original string

while i <= reverse.count {
    let indexOfString = reverse.index(reverse.startIndex, offsetBy: i)

    if reverse[indexOfString] == "1" {
        labelInputInt += 2^i * 1
    }

    i += 1
}

I am using a while loop to get the index indexOfString and check if in the string reverse at the specific index it is equal with "1".

The problem is that I get a runtime error when the if statement is executed.

The error looks like this:

2  libpthread.so.0 0x00007fc22f163390
3  libswiftCore.so 0x00007fc22afa88a0 _T0s18_fatalErrorMessages5NeverOs12Stati
cStringV_A2E4fileSu4lines6UInt32V5flagstFTfq4nnddn_n + 96
4  libswiftCore.so 0x00007fc22afb3323
5  libswiftCore.so 0x00007fc22afdf9a2
6  libswiftCore.so 0x00007fc22aedca19 _T0SS9subscripts9CharacterVSS5IndexVcfg 
+ 9
7  libswiftCore.so 0x00007fc22f591294 _T0SS9subscripts9CharacterVSS5IndexVcfg 
+ 74139780
8  swift           0x0000000000f2925f
9  swift           0x0000000000f2d402
10 swift           0x00000000004bf516
11 swift           0x00000000004ae461
12 swift           0x00000000004aa411
13 swift           0x0000000000465424
14 libc.so.6       0x00007fc22d88d830 __libc_start_main + 240
15 swift           0x0000000000462ce9
Stack dump:
0.    Program arguments: /home/drkameleon/swift4/usr/bin/swift -frontend -inte
rpret tmp/XfwP0oM7FJ.swift -disable-objc-interop -suppress-warnings -module-na
me XfwP0oM7FJ 
Illegal instruction (core dumped)

So, how can I get a specific character of a String and compare it with another character without getting this crash?

Martin R · Accepted Answer · 2017-12-17 15:53:51Z

4

Your approach to get a specific character from a string is actually correct, there are two other problems in your code:

The index i should run up to and excluding reverse.count. This is conveniently done with the "half-open range" operator (..<).
^ is the bitwise-xor operator, not exponentiation. Exponentiation is done with the pow() function, in your case
```
labelInputInt += Int(pow(2.0, Double(i)))
```
or with the "shift-left" operator << if the base is 2.

So this would be a working variant:

for i in 0 ..< reverse.count {
    let indexOfString = reverse.index(reverse.startIndex, offsetBy: i)
    if reverse[indexOfString] == "1" {
        labelInputInt += 1 << i
    }
    i += 1
}

But you can simply enumerate the characters of a string in reverse order instead of subscripting (which is also more efficient):

let binaryString = "10010"
var result = 0
for (i, char) in binaryString.reversed().enumerated() {
    if char == "1" {
        result += 1 << i
    }
}
print(result)

Even simpler with forward iteration, no reversed() or << needed:

let binaryString = "10010"
var result = 0
for char in binaryString {
    result = 2 * result
    if char == "1" {
        result += 1
    }
}
print(result)

Which suggests to use reduce():

let binaryString = "10010"
let result = binaryString.reduce(0) { 2 * $0 + ($1 == "1" ? 1 : 0) }
print(result)

But why reinvent the wheel? Just use init?(_:radix:) from the Swift standard library (with error-checking for free):

let binaryString = "10010"
if let result = Int(binaryString, radix: 2) {
    print(result)
} else {
    print("invalid input")
}

edited Dec 17, 2017 at 15:53

answered Dec 17, 2017 at 13:58

Martin R

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2 Comments

vacawama Over a year ago

(1 << i) instead of Int(pow(2.0, Double(i))) perhaps?

Martin R Over a year ago

@vacawama: You are completely right. I wanted to provide the "correct" version of exponentiation, both versions added now.

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