Use break {label} to break out of any loop as nested as you want. Just put the label before the for loop you want to break out of. This is fairly similar to the code that does a goto {label} but I think a tad more elegant, but matter of opinion I guess. package main … Read more
Swift allows for labeled statements. Using a labeled statement, you can specify which which control structure you want to break from no matter how deeply you nest your loops (although, generally, less nesting is better from a readability standpoint). This also works for continue. Example: outerLoop: while someCondition { if someOtherCondition { switch (someValue) { … Read more
I played around with domi27’s idea and came up with this. I made a nested array as my tree, [‘link’][‘sublinks’] is null or another array of more of the same. Templates The sub-template file to recurse with: <!–includes/menu-links.html–> {% for link in links %} <li> <a href=”https://stackoverflow.com/questions/8326482/{{ link.href }}”>{{ link.name }}</a> {% if link.sublinks %} … Read more
To get access to index, you need to convert your list to a map using the asMap operator. Example final fruitList = [‘apple’, ‘orange’, ‘mango’]; final fruitMap = fruitList.asMap(); // {0: ‘apple’, 1: ‘orange’, 2: ‘mango’} // To access ‘orange’ use the index 1. final myFruit = fruitMap[1] // ‘orange’ // To convert back to … Read more
Your problem is simple: names = {‘John’, ‘Joe’, ‘Steve’} for names = 1, 3 do print (names) end This code first declares a global variable called names. Then, you start a for loop. The for loop declares a local variable that just happens to be called names too; the fact that a variable had previously … Read more
Remove the brackets from [formArrayName]=”areas” and use only formArrayName=”areas” This, because with [ ] you are trying to bind a variable, which this is not. Also notice your submit, it should be: (ngSubmit)=”onSubmit(areasForm.value)” instead of areasForm.values.
[*] Use enumerate() like so: def draw_menu(options, selected_index): for counter, option in enumerate(options): if counter == selected_index: print ” [*] %s” % option else: print ” [ ] %s” % option options = [‘Option 0’, ‘Option 1’, ‘Option 2’, ‘Option 3’] draw_menu(options, 2) Note: You can optionally put parenthesis around counter, option, like (counter, option), … Read more
In Lua 5.2 the best workaround is to use goto: — prints odd numbers in [|1,10|] for i=1,10 do if i % 2 == 0 then goto continue end print(i) ::continue:: end This is supported in LuaJIT since version 2.0.1
This will print 1 to 5. The return@forEach acts like the keyword continue in Java, which means in this case, it still executes every loop but skips to the next iteration if the value is greater than 5. fun main(args: Array<String>) { val nums = listOf(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) … Read more
The idiomatic approach in Go is to write a for loop like this. for i := 1; i <= 10; i++ { fmt.Println(i) } There’s definitely advantages in ranges and they’re used in many other languages, but a Go design principle is to only introduce an abstraction if the benefits significantly outweigh the costs (including … Read more