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var a = 5 | |
var b = 8 | |
var c = a | |
a = b | |
b = c | |
print("a: \(a)") | |
print("b: \(b)") | |
I suggest you name your variable like this instead of using "c" or other words. It makes your code much more readable
var a = 5
var b = 8
var tempA = a
a = b
b = tempA
print("a: \(a)")
print("b: \(b)")
my answer is one line
var c = a; a = b; b = c;
var a = 5
var b = 8
var c = a
a = b
b = c
print("a = (a)")
print("b = (b)")
var a = 5
var b = 8
a = a+b
b = a-b
a = a-b
print("a = (a)")
print("b = (b)")
I did it like this xD
var a = 5
var b = 8
a=(a+a+a+a+a+a+a+a)/a
b=(b+b+b+b+b)/b
print("the value of a is (a)")
print("the value of b is (b)")
var a = 5
var b = 8
b = b-a
a = a + b
b = a - b
print(a,b)
My solution:
`func exercise() {
var a = 5
var b = 8
let c = a
a = b
b = c
print("a: \(a)")
print("b: \(b)")
}`
func exercise() {
var a = 5
var b = 8
var c = a
a = b
b = c
print("a: \(a)")
print("b: \(b)")
}
also nice to know how to do it without a 3rd variable:
a=a+b
b=a-b
a=a-b
This would not work for strings now. Would it?
var a = 1
var b = 2
var c = a
a = b
b = c
print("a: (a)")
print("b: (b)")
var a = 7
var b = 9
var c = a
b = a
a = b
print("a is = (a)")
print ("b is = (b)")
var a = 7
var b = 8
var c = a
a = b
b = c
func exercise() {
var a = 5
var b = 8
//Write your code here.
//Dont change any of the existing code.
print("a: \(b)")
print("b: \(a)")
}
it is working:))
var a = 5
var b = 7
a = a + b // 12
b = a - b // 12 - 7 = 5
a = a - b // 12 - 5 = 7
print("a: (a)")
print("b: (b)")
var a = 5
var b = 8
var c = b
b = a
a = c
//Replace this line with your code.
print("a: (b)")
print("b: (a)")