num += 1 is rather equivalent to ++num. All those expressions (num += 1, num++ and ++num) increment the value of num by one, but the value of num++ is the value num had before it got incremented. Illustration: int a = 0; int b = a++; // now b == 0 and a == … Read more
The typical answer to this question, unfortunately posted here already, is that one does the increment “before” remaining operations and the other does the increment “after” remaining operations. Though that intuitively gets the idea across, that statement is on the face of it completely wrong. The sequence of events in time is extremely well-defined in … Read more
An alternative is: my_dict[key] = my_dict.get(key, 0) + num
This is what I read and could answer your question: “preincrement (++i) adds one to the value of i, then returns i; in contrast, i++ returns i then adds one to it, which in theory results in the creation of a temporary variable storing the value of i before the increment operation was applied”.
$a=1; $b=$a+$a++; var_dump($b); // int(3) You assumed that the expression above is evaluated from left to right as follows (temporary variables $u and $v are introduced in the explanation for clarity): $a = 1; $u = $a; // ($a) the LHS operand of `+` $v = $a; // \ ($a++) the RHS operand of `+` … Read more
The write is atomic but an increment also requires a read. So the question is: Are you sure the read is safe, in other words, are you sure another thread doing the increment will not end up with the same value to be incremented? I have doubts. The 100% correct way of doing this would … Read more
UPDATE query places an update lock on the pages or records it reads. When a decision is made whether to update the record, the lock is either lifted or promoted to the exclusive lock. This means that in this scenario: s1: read counter for image_id=15, get 0, store in temp1 s2: read counter for image_id=15, … Read more
Other people have tackled the functional difference between post and pre increment. As far as being an lvalue is concerned, i++ can’t be assigned to because it doesn’t refer to a variable. It refers to a calculated value. In terms of assignment, both of the following make no sense in the same sort of way: … Read more
function increase_brightness(hex, percent){ // strip the leading # if it’s there hex = hex.replace(/^\s*#|\s*$/g, ”); // convert 3 char codes –> 6, e.g. `E0F` –> `EE00FF` if(hex.length == 3){ hex = hex.replace(/(.)/g, ‘$1$1’); } var r = parseInt(hex.substr(0, 2), 16), g = parseInt(hex.substr(2, 2), 16), b = parseInt(hex.substr(4, 2), 16); return ‘#’ + ((0|(1<<8) + … Read more
i++ and (i)++ behave identically. C 2018 6.5.1 5 says: A parenthesized expression is a primary expression. Its type and value are identical to those of the unparenthesized expression. It is an lvalue, a function designator, or a void expression if the unparenthesized expression is, respectively, an lvalue, a function designator, or a void expression. … Read more