increment

I'm working on an Excel spreadsheet, and I have to use only one type of formula for a huge amount of data. Since in the formula the only necessary changes concern letters, I was wondering if there is a way to make a program that increments them following the Excel columns order (A, B, C...Z; AA, AB, AC...AZ; BA, BB, BC...BZ). In my case, I would need to increment letters each time by five, so here is the kind of code I'm trying to obtain: #include <iostream> using namespace std; int main() { char x = 'B'; char y = 'E'; for (int z = 1; z < 2255; z++) { cout << "=SUMPRODUCT(SUBTOTAL(4,OFFSET(" <

I'm very new to Javascript, so I assume this is a stupid mistake. function upvote() { var score = parseInt(document.getElementById('voteScore').innerHTML); score = score++; document.getElementById('voteScore').innerHTML = score; } The div named "voteScore" contains the number 46 only (no HTML or anything). I am attempting to grab the string, convert it to an int, increment it and put it back in the div. score++ increments score , you don't need to assign it back to score . Either remove the score = or change score++ to score+1 . 来源： https://stackoverflow.com/questions/7893753/incrementing-a

I wonder how C++ behaves in this case: char variable = 127; variable++; In this case, variable now equals to -128. However did the increment operator wrapped the value to its lower bound or did an overflow occurred? An overflow occured and results in undefined behavior . Section 5.5: Ifduring the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type , the behavior is undefined [...] The standard goes on to note that integer overflows are, in most implementations, ignored. But this doesn't represent a guarantee. Plain char

I'm having problems incrementing a counter under certain conditions. Input: <Users> <User> <id>1</id> <username>jack</username> </User> <User> <id>2</id> <username>bob</username> </User> <User> <id>3</id> <username>bob</username> </User> <User> <id>4</id> <username>jack</username> </User> </Users> Wanted Output: <Users> <User> <id>1</id> <username>jack01</username> </User> <User> <id>2</id> <username>bob01</username> </User> <User> <id>3</id> <username>bob02</username> </User> <User> <id>4</id> <username>jack02</username> </User> </Users> To accomplish this following algorithm can be used:

$letter = array(); for ($i = 'A'; $i !== 'ZZ'; $i++){ $letter[] .= $i; } print_r($letter); From above script I do a loop from A , B , C , D ... ZZ . Now, I want to make it as A , C , E , G , I ... ZZ . ( 2 steps instead of 1 ) I need direction to do it. mickmackusa ord() will not work because your end string is two characters long. Returns the ASCII value of the first character of string . Watch it break . From my testing, you need to check that the end string doesn't get "stepped over". The perl-style character incrementation is a cool method, but it is a single-stepping method. For this

I have a class Transaction which has a var amount of type Int . I want to access it from another class, where I have an array of Transactions and sum all of their amounts. So I have this piece of code func computeTotal()-> Int{ let total = 0 for transaction in transactions{ //get the amounts of each and sum all of them up total += transaction.amount } return total } But it gives me an error Cannot invoke "+=" with an argument list of type (Int, @value Int) What can cause that? I know that in Swift both operands must be the same type, but they are both of type Int in my code. let creates an

I have never seen the usecase for pre-increment and post-increment in actual code. The only place i see them most often are puzzles. My opinion is, it introduces more confusion rather than being useful. is there any real use case scenario for this can't this can be done by using += y = x++ y = x x += 1 It's just a shorter way of writing the same thing and it's only confusing to those who don't deeply understand C (a) . The same argument could be made for replacing: for (i = 0; i < 10; i++) printf ("%d\n", i); with: i = 0; while (i < 10) { printf ("%d\n", i); i = i + 1; } since any for can also

Consider the following code: for (float i = 0f; i < int.MaxValue; i++) { // Some code } Which is supposed to loop from 0 to int.MaxValue (2 31 -1), but it doesn't. Once i reached 2 24 , i++ doesn't work anymore for a reason that I'm totally unable to understand. In the Immediate Window of VS I've try this: >i 16777216.0 >i + 1 16777216.0 // ??? >i == i + 1 false // as expected, but a lack of consistency with upper statement >i + 2 16777218.0 Why does it behave like so? What is special with 2 24 +1? Hamid Pourjam It is because of float precision. It is based on IEEE Standard for Floating-Point