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| 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
print("The value of a is (b)")
print("The value of b is (a)")
Alex. id you mean what I put above? That is what first came to mind to solve it but it was not using 3 lines of code. It is was clear that we were suppose to write 3 lines of code but I was not clear why in the original probalem those print statements were there or what to take out and what to leave in.
var a = 5
var b = 8
var c = 3
print("a: (a+c)")
print("b: (b-c)")
I guess I cheated a lot lol
A bit longer but still :)
var c = b - a
a = b
b -= c
func swap(a:inout Int, b:inout Int){
a = a + b
b = a - b
a = a - b
}
var a = 5
var b = 8
swap(&a, &b)
print("a: (a)")
print("b: (b)")
var a = 5
var b = 8
print("a: (b)")
print("b: (a)")
/*
print("a: (a)")
print("b: (b)")
*/
var c = a
a = b
b = c
print("a: (a)")
print("b: (b)")
var a = 5
var b = 8
a = a + b // a=13
b = a - b //b=5
a = a - b //a=8
print("a: (a)")
print("b: (b)")
// Checkout my code as below keep it simple
var a = 5
var b = 8
var c = a + b
//Replace this line with your code.
a = c - a
b = c - b
print("a: (a)")
print("b: (b)")
please tell i did it. is it correct
by entering this code between variables and print statement
i entered this :
a = b
b = a
and it worked perfectly
Can you please share your complete code @krrish-cmd
Use in between print will work but at what point you changed the Variable need to check.
please tell i did it. is it correct
by entering this code between variables and print statement
i entered this :
a = b
b = aand it worked perfectly
var a = 5
var b = 8
if you assign a = b that will be 8 and then
again the b which is holding new value 8, you assign it to a,
that means both of these links print the same i.e. 8.
var a = 5
var b = 8
(a, b) = (b, a)
var a = 5
var b = 8
(a, b) = (b, a)
print("The value of a is: (a)")
print("the value of b is: (b)")
Output:
The value of a is: 8
the value of b is: 5
var a = 5
var b = 8
a += 3
b -= 3
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)")
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 = 8
print("The value of a is (a)")
print("The value of a is (b)")
I was kind of confused about what to do with the code and how to manage it without adding a third variable. 😟