You’re reseeding the pseudo-random sequence with the same seed before each call to rand, since the value returned by time only changes once a second. That causes you to get the same result each time. Only call srand once, at the start of the program. Better still, look at the C++11 <random> library, which has … Read more
The values of your variables throughout the loops are like follows: x y result ———- 0 0 1 0 1 2 0 2 3 1 0 4 1 1 5 1 2 6 2 0 7 2 1 8 2 2 9 3 0 10 3 1 11 3 2 12 So that’s why result … Read more
You can and probably should amortize the runtime overhead of creating and destroying an OpenMP parallel region by putting all of your parallel constructions – in this case, two do loops and a pair of sections – within a single parallel region. The following code compiles and executes as expected. program main implicit none integer … Read more
I’ll take one stab at this with two implementations. First, I’ll use a character vector. If yours is in a frame, replace it with myframe$mycolumn. v <- c(“110231 validation 108871 validation 85933”, “21102 validation 93442 21232 validation 73769 26402 validation 127221 26402”, “99763 99763 validation 99763 validation 99763”, “validation 199022 validation 122099 validation 12209 validation … Read more
This should work: <?php $re=”/(\d{1,})/m”; $str=”<b>100£</b> was £160″; preg_match_all($re, $str, $matches, PREG_SET_ORDER, 0); if(count($matches)>0 && isset($matches[0][0]) && isset($matches[1][0])) { echo(sprintf(‘<b>£%d</b> was £%d’, $matches[0][0], $matches[1][0])); } 0 solved Move Pound £ symbol from after a number to before it
You can pair adjacent numbers, calculate the differences between the pairs, and determine that the list forms an arithmetic progression if the number of unique differences is no greater than 1: from operator import sub def progression(l): return len(set(sub(*p) for p in zip(l, l[1:]))) <= 1 so that: print(progression([3])) print(progression([7,3,-1,-5])) print(progression([3,5,7,9,10])) outputs: True True False … Read more
I know an answer has been accepted, but it doesn’t actually satisfy the requirements fully, i.e. Each callback should take a single, integer value as a parameter which is the amount to increment the counter’s existing value by. The accepted answer takes the event object as a parameter which is not the specified requirement. The … Read more
package main import “fmt” func main() { //create a map m := map[int64]int64{512: 8, 513: 9, 234: 9, 392: 0} //create a slice to hold required values s := make([][]int64, 0) //range over map `m` to append to slice `s` for k, v := range m { // append each element, with a new slice … Read more
We don’t have the full details on your string class, so this is hard to reproduce. However, when you call sensors[0] = sensor1; you are copying your Sensor, but haven’t defined an assignment operator (or copy constructor for that matter). You also do this in the Car constructor with (*(m_sensors + i)) = Sensor(*(sensors + … Read more
You can try with String.prototype.slice() The slice() method extracts a section of a string and returns it as a new string, without modifying the original string. and String.prototype.lastIndexOf() The lastIndexOf() method returns the index within the calling String object of the last occurrence of the specified value, searching backwards from fromIndex. Returns -1 if the … Read more
From the documentation: Indentation is rejected as inconsistent if a source file mixes tabs and spaces in a way that makes the meaning dependent on the worth of a tab in spaces; a TabError is raised in that case. You state the body of the with statement is indented with 3 tabs. If that is … Read more
In GeeksForGeeks, overloading is defined to be “a feature in C++ where two or more functions can have the same name but different parameters.” Constructor can be overloaded by allowing for same names and different parameters. For example, you can refer to this. Nesting functions is not supported by C++. Please refer to this link. … Read more
You can use paste to paste the day with the ‘Date’ column, convert to Date class using as.Date with the appropriate format argument. df1$Date <- as.Date(paste(df1$Date, ’01’), ‘%Y %B %d’) For more info, you can check ?as.Date, ?strptime, ?as.POSIXct 1 solved Converting a year-month column to date type [duplicate]
the problem turned out to be the max char size that can be passed to text to speech engine is about 4000 if your string contains more than 4000 characters it will not work // this will not work tts.speak(text, TextToSpeech.QUEUE_ADD, null); in order to get it working add an setOnUtteranceProgressListener and loop trough your … Read more
You can try this: Get-WmiObject -Class win32_Logicaldisk -ComputerName localhost | where {($_.DriveType -eq 3 -or $_.DriveType -eq 2) -and $_.FileSystem -ne $null } | Select -Property @{Name=”Volume”;Expression = {$_.DeviceID -replace “:”,””}}, @{Name=”Free”;Expression = { “{0:N0}%” -f (($_.FreeSpace/$_.Size) * 100) } } solved power shell script to get drive space ( percentage used and free ) … Read more