var x = ["a", "b", "c"];
for(var i = 0; i < x.length; i++){
x[i] = x[2 - i];
}
My approach:
for i = 0 => x[0] = x[2] (which is "c", so replace "a" with "c")
for i = 1 => x[1] = x[1] (which is "b", so replace "b" with "b")
for i = 2 => x[2] = x[0] (which is "a" so replace "c" with "a")
for i = 3 test failed, stop.
so x = ["c", "b", "a"]
Why does the console return x as ["c","b","c"]? Could somebody please tell me whether I have completely misunderstood loop logic? Thank you!
var x = ["a", "b", "c"];
for(var i = 0; i < x.length; i++){
x[i] = x[2 - i];
}
Let's write this code out longhand:
var x = ['a', 'b', 'c'];
x[0] = x[2]; // ['c', 'b', 'c']
x[1] = x[1]; // ['c', 'b', 'c']
x[2] = x[0]; // ['c', 'b', 'c']
The problem is that by the time you get to i = 2 you've already modified x[0] = x[2], so x[2] = x[0] unsurprisingly has no result.
You can use the Array#reverse method, I think:
var x = ['a', 'b', 'c'];
x.reverse(); // ['c', 'b', 'a']
edit: Neat! Didn't know about Array.reverse(), that's definitely easier than below!
By the time the third iteration occurs, the first element has already been set to "c" in the first iteration.
The easiest way to do it is to simply make a second array for output:
var x = ["a", "b", "c"];
var y = new Array(x.length);
for(var i = 0; i < x.length; i++){
y[i] = x[2 - i];
}
console.log(y)
When i = 2, x[0] is "c" , you replaced it when "i" was 0. So it will be ["c","b","c"].
x.reverse()?