In your particular example, none because the compiler internally uses StringBuilders to do String concatenation. If the concatenation occurred in a loop, however, the compiler could create several StringBuilder and String objects. For example: String s= “” ; for(int i= 0 ; i < 10 ; i++ ) s+= “a” ; Each time line 3 … Read more
Modern string formatting: “{} and {}”.format(“string”, 1)
Concatenation is (significantly) faster according to my machine. But stylistically, I’m willing to pay the price of substitution if performance is not critical. Well, and if I need formatting, there’s no need to even ask the question… there’s no option but to use interpolation/templating. >>> import timeit >>> def so_q_sub(n): … return “%s%s/%d” % (DOMAIN, … Read more
There should be more elegant solutions, but simplest one I can think of would be this. // The data you have pasted in the question $data = []; $groups = []; // Go through the entire array $data foreach($data as $item){ // If the key doesn’t exist in the new array yet, add it if(!array_key_exists($item[1], … Read more
You can do that in several ways: As you shown with << but that is not the usual way With string interpolation source = “#{ROOT_DIR}/#{project}/App.config” with + source = “#{ROOT_DIR}/” + project + “/App.config” The second method seems to be more efficient in term of memory/speed from what I’ve seen (not measured though). All three … Read more
If you’re already on EL 3.0 (Java EE 7; WildFly, Tomcat 8, GlassFish 4, etc), then you could use the new += operator for this: <h:commandButton … action=”#{someController.doSomething(id += ‘SomeTableId’)}” /> If you’re however not on EL 3.0 yet, and the left hand is a genuine java.lang.String instance (and thus not e.g. java.lang.Long), then use … Read more
Put your values in an array and then implode that array with a comma (+ a space to be cleaner): $myArray = array(); foreach ($this->sinonimo as $s){ $myArray[] = ‘<span>’.ucfirst($s->sinonimo).'</span>’; } echo implode( ‘, ‘, $myArray ); This will put commas inbetween each value, but not at the end. Also in this case the comma … Read more
You found one answer, i.e. eval(parse()) . You can also investigate do.call() which is often simpler to implement. Keep in mind the useful as.name() tool as well, for converting strings to variable names.
You can do it with paste function > paste(c(‘A’, ‘B’, ‘C’), collapse=”, ” ) [1] “A, B, C”
Try this: static void Main(string[] args) { int partLength = 35; string sentence = “Silver badges are awarded for longer term goals. Silver badges are uncommon.”; string[] words = sentence.Split(‘ ‘); var parts = new Dictionary<int, string>(); string part = string.Empty; int partCounter = 0; foreach (var word in words) { if (part.Length + word.Length … Read more