I have this code:
count = $content.find('.post').length;
for x in [1...count]
/*
prev_el_height += $("#content .post:nth-child(" + x + ")").height();
*/
prev_el_height += $content.find(".post:nth-child(" + x + ")").height();
I expected this to turn into
for (x = 1; x < count; x++) { prev_el ... }
but it turns into this:
for (x = 1; 1 <= count ? x < count : x > count; 1 <= count ? x++ : x--) {
Can somebody please explain why?
EDIT: How do I get my expected syntax to output?
In CoffeeScript, you need to use the by keyword to specify the step of a loop. In your case:
for x in [1...count] by 1
...
You're asking to loop from 1 to count, but you're assuming that count will always be greater-than-or-equal-to one; the generated code doesn't make that assumption.
So if count is >= 1 then the loop counter is incremented each time:
for (x = 1; x < count; x++) { /* ... */ }
But if count is < 1 then the loop counter is decremented each time:
for (x = 1; x > count; x--) { /* ... */ }
Well, you want x to go from 1 to count. The code is checking whether count is bigger or smaller than 1.
If count is bigger than 1, then it has to increment x while it is smaller than count.
If count is smaller than 1, then it has to decrement x while it is bigger than count.
For future reference:
$('#content .post').each ->
prev_el_height += $(this).height()
Has the same effect, assuming :nth-child is equivalent to .eq(), and x going past the number the elements is a typo.
Related
i just started learning dart but there is something i cant figure out.
the first loop it prints me from 1 3 5 7 9
the second one it prints for me from 0 to 9.
why did it remove the even numbers from the loop below? i only added a variable in the first loop
void main () {
for(double a = 0; a <10 ; a++)
{
double b = a++;
print (a);
}
print("---");
for(double a = 0; a <10 ; a++)
{
print(a);
}
}
the a++ is a shortcut to a = a +1, which means that b in each step is getting the value of a+1 and the second loop would only print odd numbers since you are jumping 2 steps in each loop (a++ in the loop brackets and the b = a++)
I'm having a great deal of difficulty translating the following for loop into Swft
for (int i = 0; i * denomAmount <= amount; i++ {
}
my attempt is
for i in stride(from: 0, to: amount, by: denom){
}
but obviously it amount is not the maximum value of i. So what is the best way to approach this?
Most likely either demonAmount or amount is being updated inside the loop. So I would use a while loop:
var i = 0
while i * demonAmount <= amount {
// the loop code
i += 1
}
I want to know how am I supposed to count the number of time a loop has repeated itself? More specifically how do I extract and output the number of repeats?
var x = 20
while x < 100 {
x += 10
}
The loop has executed 8 times in order to get x == 100. Is there a way to extract the number '8' so it can be used somewhere else (e.g. to make it a variable elsewhere)?
You said it yourself: you want to count. So count!
var x = 20
var numtimes = 0
while x < 100 {
x += 10
numtimes += 1 // count!
}
numtimes // 8
I have written a program that generates prime numbers . It works well but I want to speed it up as it takes quite a while for generating the all the prime numbers till 10000
var list = [2,3]
var limitation = 10000
var flag = true
var tmp = 0
for (var count = 4 ; count <= limitation ; count += 1 ){
while(flag && tmp <= list.count - 1){
if (count % list[tmp] == 0){
flag = false
}else if ( count % list[tmp] != 0 && tmp != list.count - 1 ){
tmp += 1
}else if ( count % list[tmp] != 0 && tmp == list.count - 1 ){
list.append(count)
}
}
flag = true
tmp = 0
}
print(list)
Two simple improvements that will make it fast up through 100,000 and maybe 1,000,000.
All primes except 2 are odd
Start the loop at 5 and increment by 2 each time. This isn't going to speed it up a lot because you are finding the counter example on the first try, but it's still a very typical improvement.
Only search through the square root of the value you are testing
The square root is the point at which a you half the factor space, i.e. any factor less than the square root is paired with a factor above the square root, so you only have to check above or below it. There are far fewer numbers below the square root, so you should check the only the values less than or equal to the square root.
Take 10,000 for example. The square root is 100. For this you only have to look at values less than the square root, which in terms of primes is roughly 25 values instead of over 1000 checks for all primes less than 10,000.
Doing it even faster
Try another method altogether, like a sieve. These methods are much faster but have a higher memory overhead.
In addition to what Nick already explained, you can also easily take advantage of the following property: all primes greater than 3 are congruent to 1 or -1 mod 6.
Because you've already included 2 and 3 in your initial list, you can therefore start with count = 6, test count - 1 and count + 1 and increment by 6 each time.
Below is my first attempt ever at Swift, so pardon the syntax which is probably far from optimal.
var list = [2,3]
var limitation = 10000
var flag = true
var tmp = 0
var max = 0
for(var count = 6 ; count <= limitation ; count += 6) {
for(var d = -1; d <= 1; d += 2) {
max = Int(floor(sqrt(Double(count + d))))
for(flag = true, tmp = 0; flag && list[tmp] <= max; tmp++) {
if((count + d) % list[tmp] == 0) {
flag = false
}
}
if(flag) {
list.append(count + d)
}
}
}
print(list)
I've tested the above code on iswift.org/playground with limitation = 10,000, 100,000 and 1,000,000.
On coffeescript loop 'for'
eg.
if 1 < x, code like below:
console.debug i for i in [1..0]
Generated code is:
var i;
for (i = 1; i >= 0; i--) {
console.debug(i);
}
if 1 > x,code like below:
console.debug i for i in [1..2]
Generated code is:
var i;
for (i = 1; i <= 2; i++) {
console.debug(i);
}
If i want write that javascript.How to ?
for(var i=1;i<=0;i++){
console.debug(i);
}
Because i don't know the condition is greater than left side or less than left side.
But i just want it i++
What's wrong with me?
EDIT BELOW:
For coffeescript's feature,I add condition before the loop or add condition on for loop.
eg:
if x - y >=1
console.debug i for i in [1..x-y]
or
console.debug i for i in [1..x-y] and x-y >=1
That's my way.Some one have good advice?
It looks like you want to do this:
console.debug i for i in [1..x-y] by 1
Which gets compiled to:
var i, _i, _ref;
for (i = _i = 1, _ref = x - y; _i <= _ref; i = _i += 1) {
console.debug(i);
}
for(var i=1;i<=0;i++){
console.debug(i);
}
is equivalent to
var i = 1;
while(true) {
console.debug(i);
i++;
}
which in coffeescript is written as
i = 1
while true
console.debug(i);
i++;