modulus

Detect each 4 using modulus php

狂风中的少年 提交于 2019-12-01 23:59:14
I am trying to detect each 4th post to insert extra code in my layout in wordpress using modulus method but I just cant get it. Here is a short example of mine: <?php if (have_posts()) : ?> <?php $count=0;?> <?php while (have_posts()) : the_post(); ?> <div class="column"> <!--content--> </div> <?php if ($count % 4 == 0){ echo '<div class="clear"></div>'; } $count++; ?> <?php endwhile; ?> <?php endif; ?> all that is inside the while loop. What am I doing wrong? Thank you. You need to start your counter at 1, as you are increasing it at the end of the loop: <?php $count=1;?> Either that, or you

Why does Java's % operator give different results than my calculator for a negative dividend?

两盒软妹~` 提交于 2019-12-01 20:08:29
问题 How come on a calculator -1 mod 26 = 25, but in C or Java -1 % 26 == -1 . I need a program which solves it like the calculator. Is there a difference between the two? 回答1: Both answers (25 and -1) are valid. It's just that different systems have different conventions. The one I see the most common (in mathematics) is: quotient = floor(x / y) remainder = x - quotient * y Where floor() is to round towards negative infinity. This is the convention that your calculator is giving you. (Mathematica

modulus operator to run 1st and then every 3rd item

一曲冷凌霜 提交于 2019-12-01 18:01:52
So i need it to run on the first loop and then every 3rd loop if ($k % 3 || $k==1 ) { echo '<div class="modcontainer">'; } Seems simple to me, but i don't have the understanding of modulus Modulus returns the remainder, not a boolean value. This code will resolve to true for 1, 3, 6, 9, ... if (($k % 3 == 0) || $k==1 ) { echo '<div class="modcontainer">'; } This code will resolve to true for 1, 4, 7, 10, ... if ($k % 3 == 1) { echo '<div class="modcontainer">'; } 来源: https://stackoverflow.com/questions/11261192/modulus-operator-to-run-1st-and-then-every-3rd-item

modulus operator to run 1st and then every 3rd item

空扰寡人 提交于 2019-12-01 17:18:49
问题 So i need it to run on the first loop and then every 3rd loop if ($k % 3 || $k==1 ) { echo '<div class="modcontainer">'; } Seems simple to me, but i don't have the understanding of modulus 回答1: Modulus returns the remainder, not a boolean value. This code will resolve to true for 1, 3, 6, 9, ... if (($k % 3 == 0) || $k==1 ) { echo '<div class="modcontainer">'; } This code will resolve to true for 1, 4, 7, 10, ... if ($k % 3 == 1) { echo '<div class="modcontainer">'; } 来源: https:/

`java (0 % 2 != 0) == false`

巧了我就是萌 提交于 2019-12-01 16:34:57
The part I keep getting stuck on is boolean(0 % 2 !=0) == false. I mean if 2 goes into 0, 0 times then the remainder would be 2, and 2 does not equal 0. So it should be true. Yet but when I put the boolean in my java program it is treating it as false. Anyone know why? The only logical answer I can wrap my head around is that maybe integers go into 0 and infinite number of times and so are recognized as false, anyone? There are two steps: 0 % 2 evaluates to 0 . 0 != 0 evaluates to false . To elaborate on the first step, the JLS defines the % operator like so: The binary % operator is said to

Python Modulus Giving String Formatting Errors

女生的网名这么多〃 提交于 2019-12-01 16:20:51
问题 I'm trying to perform the modulus of a value in python, but I'm getting errors as it's interpretting the modulus as a string formatting constant, from my knowledge. My initial guess would be to type cast this, but then it hangs. val = pow(a,n,p) val = y1*val val = val % p Are the two lines of code corresponding to this question. Right now, when I run this, I get: TypeError: not all arguments converted during string formatting At the second line. If I wrap val into an integer and type cast it.

Is there a reason some languages allow a negative modulus?

给你一囗甜甜゛ 提交于 2019-12-01 15:55:50
I am curious about these languages (Java, C ...) which ignore mathematical definition of modulus operation. What is the point of returning negative values in a module operation (that, by definition, should allways return a positive number)? I doubt that the remainder operator was deliberately designed to have those semantics, which I agree aren't very useful. (Would you ever write a calendar program that shows the weekdays Sunday, Anti-Saturday, Anti-Friday, ..., Anti-Monday for dates before the epoch?) Rather, negative remainders are a side effect of the way integer division is defined. A rem

Is there a reason some languages allow a negative modulus?

元气小坏坏 提交于 2019-12-01 15:40:19
问题 I am curious about these languages (Java, C ...) which ignore mathematical definition of modulus operation. What is the point of returning negative values in a module operation (that, by definition, should allways return a positive number)? 回答1: I doubt that the remainder operator was deliberately designed to have those semantics, which I agree aren't very useful. (Would you ever write a calendar program that shows the weekdays Sunday, Anti-Saturday, Anti-Friday, ..., Anti-Monday for dates

What is the modulo operator for longs in Java?

非 Y 不嫁゛ 提交于 2019-12-01 02:44:47
How do I find the modulo (%) of two long values in Java? My code says 'Integer number too large' followed by the number I'm trying to mod. I tried casting it to a long but it didn't work. Do I have to convert it to a BigInteger and use the remainder method? Thanks. The % operator does work for longs. It sounds like you may have forgotten to stick L at the end of a numeric literal, as in 123456789L . Can we see your code? You can only have an integer up to 2 147 483 647. If you want to go bigger than that, say 3 billion, you must specify it to be a long class Descartes { public static void main

C : Modulus operator on unsigned int gives unexpected output

旧城冷巷雨未停 提交于 2019-11-30 09:12:03
问题 #include <stdio.h> main() { unsigned a = -20; unsigned b = 10; printf("%d\n", (a % b)); printf("%d\n", (-20 % 10)); } Output: 6 0 The second printf prints the expected value of 0 while the first printf prints 6. Why this unexpected output with unsigned ints? 回答1: unsigned int can hold values from 0 to UINT_MAX , no negative values. So -20 is converted to -20 + UINT_MAX + 1 . On your system: (-20 + UINT_MAX + 1) % 10 != -20 % 10 回答2: And what are you expecting? a % b is equivalent to, let's