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

Answer 1

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")
}