We can break both a $(selector).each() loop and a $.each() loop at a particular iteration by making the callback function return false. Returning non-false is the same as a continue statement in a for loop; it will skip immediately to the next iteration. return false; // this is equivalent of ‘break’ for jQuery loop return; … Read more
The first form is efficient only if vector.size() is a fast operation. This is true for vectors, but not for lists, for example. Also, what are you planning to do within the body of the loop? If you plan on accessing the elements as in T elem = some_vector[i]; then you’re making the assumption that … Read more
Most of the times it is easier (and cheaper) to make the first iteration the special case instead of the last one: first = True for data in data_list: if first: first = False else: between_items() item() This will work for any iterable, even for those that have no len(): file = open(‘/path/to/file’) for line … Read more
Looping through the characters of a std::string, using a range-based for loop (it’s from C++11, already supported in recent releases of GCC, clang, and the VC11 beta): std::string str = ???; for(char& c : str) { do_things_with(c); } Looping through the characters of a std::string with iterators: std::string str = ???; for(std::string::iterator it = str.begin(); … Read more
The best answer here is to use all(), which is the builtin for this situation. We combine this with a generator expression to produce the result you want cleanly and efficiently. For example: >>> items = [[1, 2, 0], [1, 2, 0], [1, 2, 0]] >>> all(flag == 0 for (_, _, flag) in items) … Read more
It’s not that i– is faster than i++. Actually, they’re both equally fast. What takes time in ascending loops is evaluating, for each i, the size of your array. In this loop: for(var i = array.length; i–;) You evaluate .length only once, when you declare i, whereas for this loop for(var i = 1; i … Read more
You can use a range with a step size of 2: Python 2 for i in xrange(0,10,2): print(i) Python 3 for i in range(0,10,2): print(i) Note: Use xrange in Python 2 instead of range because it is more efficient as it generates an iterable object, and not the whole list.
No – but technically there is a work-around (not that i’d actually use it unless forced to): for(struct { int a; char b; } s = { 0, ‘a’ } ; s.a < 5 ; ++s.a) { std::cout << s.a << ” ” << s.b << std::endl; }
You have three (or so) options to break out of loops. Suppose you want to sum numbers until the total is greater than 1000. You try var sum = 0 for (i <- 0 to 1000) sum += i except you want to stop when (sum > 1000). What to do? There are several options. … Read more
In addition to the solutions provided by @Audi, there’s also forEachIndexed: collection.forEachIndexed { index, element -> // … }