Swift If Or/And Statement Like Python

swift if or/and statement like python

You can use

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&&

for logical and

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|| 

for logical or

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so you can do

if a > 0 && i == j || f < 3 {
    ...
}

see here
https://developer.apple.com/library/ios/documentation/swift/conceptual/Swift_Programming_Language/BasicOperators.html

Python style conditional expression in Swift 3

Great question :)

Rather than storing the expression, store the result as an optional if the conditional is false. Edited based on some comments, resulting in some much cleaner code.

infix operator ==| : TernaryPrecedence // if/where
infix operator |!= : TernaryPrecedence // else

func ==|<T> (lhs: @autoclosure () -> T, rhs: T?) -> T {
    return rhs ?? lhs()
}

func |!=<T> (lhs: Bool, rhs: @autoclosure () -> T) -> T? {
    return lhs ? nil : rhs()
}

Swift if statement - multiple conditions separated by commas?

Yes when you write

if let a = optA, let b = optB, let c = optC {
    
}

Swift does execute the body of the IF only if all the optional bindings are properly completed.

More

Another feature of this technique: the assignments are done in order.

So only if a value is properly assigned to a, Swift tries to assign a value to b. And so on.

This allows you to use the previous defined variable/constant like this

if let a = optA, let b = a.optB {

}

In this case (in second assignment) we are safely using a because we know that if that code is executed, then a has been populated with a valid value.

If an if statement gets called and all the conditions are true, do all the else if statements get called also?

No, and you can visualize it like this:

if coins > 19 && speedLvl == 1 {
    speedLvl = 2
    coins = coins - 20
}
else {
    if coins > 49 && speedLvl == 2 {
        speedLvl = 3
        coins = coins - 50
    }
    else {
        if coins > 99 && speedLvl == 3 {
            speedLvl = 4
            coins = coins - 100
        }
    }
}

Although this code would be more easily written in Swift 4 as:

switch (speedLvl, coins) {
case (1, 20..<50):
    speedLvl += 1
    coins -= 20

case (2, 50..<100): 
    speedLvl += 1
    coins -= 50 

case (3, 100...):
    speedLvl += 1
    coins -= 100

default: break;
}

or better yet, perhaps:

let levelUpCosts = [0, 20, 50, 100]

let levelUpCost = levelUpCosts[speedLvl]
if levelUpCost < coins {
    coins -= levelUpCost
    speedLvl += 1
}

If you want to multiple level ups to be possible, all in one shot, then you can do something like this:

let levelUpCosts = [0, 20, 50, 100]

var affordedLevelUpsCost = 0
let affordedLevelUps = levelUpCosts.lazy.prefix(while: { cost in
    let newCost = affordedLevelUpsCost + cost
    let canAffordLevelUp = newCost < coins
    if canAffordLevelUp { affordedLevelUpsCost = newCost }
    return canAffordLevelUp
})

speedLvl += affordedLevelUps.count
coins -= affordedLevelUpsCost

Can I use the OR operator or AND operator between two if..else statement?

Combine them into one statement like this:

if age < 30 AND height_feet > 6.1 then
    ' do something
else
    ' do something else
end if

If you have to check the OR condition, just switch the AND to OR.

EDIT

You can combine if statements like this also:

if age < 30 AND height_feet > 6.1 then
    ' do something
else if age < 30 OR height_feet > 6.1 then
    ' do something
else
    if employee_type = 1 then 
        ' do something
    else
        ' do something
    end if
end if

Swift equivalent for while loops with else in python

There is no while ... else ... in Swift, you will have to refactor your logic.

Also, you do not want to loop inside touchesBegan your app will stop responding!

Can I use the range operator with if statement in Swift?

You can use the "pattern-match" operator ~=:

if 200 ... 299 ~= statusCode {
    print("success")
}

Or a switch-statement with an expression pattern (which uses the pattern-match
operator internally):

switch statusCode {
case 200 ... 299:
    print("success")
default:
    print("failure")
}

Note that ..< denotes a range that omits the upper value, so you probably want
200 ... 299 or 200 ..< 300.

Additional information: When the above code is compiled in Xcode 6.3 with
optimizations switch on, then for the test

if 200 ... 299 ~= statusCode

actually no function call is generated at all, only three assembly instruction:

addq    $-200, %rdi
cmpq    $99, %rdi
ja  LBB0_1

this is exactly the same assembly code that is generated for

if statusCode >= 200 && statusCode <= 299

You can verify that with

xcrun -sdk macosx swiftc -O -emit-assembly main.swift

As of Swift 2, this can be written as

if case 200 ... 299 = statusCode {
    print("success")
}

using the newly introduced pattern-matching for if-statements.
See also Swift 2 - Pattern matching in "if".

How does Swift handle if statements?

Yes

When you concatenate a list of conditions C[0]...C[n] with the AND && operator, the runtime evaluates in order each condition and if a C[i] condition is found false, then the evaluation of the whole expression does end and it is judged false.

let c0 = true
let c1 = false
let c2 = true

if c0 && c1 && c2 {
    print("Hello world")
}

In this case only c0 and c1 will be evaluated and the whole expression will be interpreted as false.

You can test it yourself in Playground.

c0 || c1 || c2

Symmetrically if you define an expression as the OR || concatenation of several clauses, then the whole expression is interpreted as true (and the evaluation of the clauses does stop) as soon as the first true condition gets found.

Is there any expression in Swift that is similar to Python's for else syntax

You could just introduce a flag to record if break is called or not. The statement

for a in b:
    if c(a):
        break
else:
    d()

is the same as

found = False
for a in b:
    if c(a):
        found = True
        break
if not found:
    d()

But note that you don't need the for char in item.characters loop at all, since you could just use the Set.isDisjointWith(_:) method.

if cur_word.isDisjointWith(item.characters) {
    maximum = ...
}

(On Swift 3 this method is renamed to Set.isDisjoint(with:))

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