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By : Mikko Waris
Date : January 12 2021, 08:33 AM
will be helpful for those in need You need to use number_format(), but there's no way of telling the function that the last 3 numbers are the decimals. So you need to first divide by 1000, and then use the function: code :
``````<?php
\$n = 1003636;
\$n /= 1000;
echo number_format(\$n, 3, ",", "."); // 1.003,636 - 3 decimals, separate thousands by . and decimals by ,
`````` ## How to convert a 16-bit hexadecimal number to binary number and then count number of zeros and one in it ? Using simples

By : Randall
Date : March 29 2020, 07:55 AM
Hope that helps A simple program I thought of was to make conversion of hexadecimal number to binary number and then using a counter to count the number of zeros and ones.
code :
``````#include <stdio.h>
#define MAX 1000

int main()
{
char binarynum[MAX], hexa[MAX];
long int i = 0,number_of_zeros=0,number_of_ones=0;

printf("Enter the value for hexadecimal ");
scanf("%s", hexa);
printf("\n Equivalent binary value: ");

while(hexa[i])
{
switch (hexa[i])
{
case '0':
printf("0000");
number_of_zeros +=4;
break;
case '1':
printf("0001");
number_of_zeros +=3;
number_of_ones +=1;
break;
case '2':
printf("0010");
number_of_zeros +=3;
number_of_ones +=1;
break;
case '3':
printf("0011");
number_of_zeros +=2;
number_of_ones +=2;
break;
case '4':
printf("0100");
number_of_zeros +=3;
number_of_ones +=1;
break;
case '5':
printf("0101");
number_of_zeros +=2;
number_of_ones +=2;
break;
case '6':
printf("0110");
number_of_zeros +=2;
number_of_ones +=2;
break;
case '7':
printf("0111");
number_of_zeros +=1;
number_of_ones +=3;
break;
case '8':
printf("1000");
number_of_zeros +=3;
number_of_ones +=1;
break;
case '9':
printf("1001");
number_of_zeros +=2;
number_of_ones +=2;
break;
case 'A': case 'a':
printf("1010");
number_of_zeros +=2;
number_of_ones +=2;
break;
case 'B': case 'b':
printf("1011");
number_of_zeros +=1;
number_of_ones +=3;
break;
case 'C': case 'c':
printf("1100");
number_of_zeros +=2;
number_of_ones +=2;
break;
case 'D': case 'd':
printf("1101");
number_of_zeros +=1;
number_of_ones +=3;
break;
case 'E': case 'e':
printf("1110");
number_of_zeros +=1;
number_of_ones +=3;
break;
case 'F': case 'f':
printf("1111");
number_of_ones +=4;
break;

default:
printf("\n Invalid hexa digit %c ", hexa[i]);
return 0;
}
i++;
}
printf("\nNumber of zero\'s is %d\n",number_of_zeros);
printf("Number of One\'s is %d\n",number_of_ones);

return 0;
}
`````` ## JQ convert to number, convert to boolean when generating new json from shell variables

By : user2446539
Date : March 29 2020, 07:55 AM
To fix the issue you can do You can use test filter to do the check and return true/false. And I think you can just use the filters directly, no need to create a json and convert again.
code :
``````happystring="Bob Ross"
unhappynumber1="1942"
unhappyboolean=true

JSON=\$(jq -n \
--arg happystring "\$happystring" --arg unhappynumber1 "\$unhappynumber1" \
--arg unhappyboolean \$unhappyboolean \
'
{
happystring: \$happystring,
unhappynumber1: \$unhappynumber1 | tonumber,
unhappyboolean: \$unhappyboolean | test("true")
}
')

echo "\$JSON" | jq
``````
``````{
"happystring": "Bob Ross",
"unhappynumber1": 1942,
"unhappyboolean": true
}
`````` ## How to convert a fix length char* to a number or convert a non-null terminated char* to number?

By : bbowles
Date : March 29 2020, 07:55 AM
I hope this helps you . You can use std::from_chars and it will get you most of the way there out of the box. It takes pointers to the beginning element and one past the last element and converts it to a number. If it can't convert the string at all it will populate the error code in the return object otherwise it converts what it can and returns a pointer to the part of the data it couldn't convert in the return object. Using it in your code changes it to
code :
``````int main ()
{
const char* arrNumber = "ff000A;";

int i_first, i_second;
std::from_chars(arrNumber, arrNumber + 2, i_first, 16);
std::from_chars(arrNumber + 2, arrNumber + 6, i_second, 16);

std::cout << "i_first: " << i_first << std::endl;
std::cout << "i_second: " << i_second << std::endl;
return 0;
}
``````
``````template<typename T>
auto from_chars(char const * begin, char const * end, int base = 10)
{
T ret{};
std::from_chars(begin, end, ret, base);
return ret;
}
``````
``````int main ()
{
const char* arrNumber = "ff000A;";

const int i_first = from_chars<int>(arrNumber, arrNumber + 2, 16);
const int i_second = from_chars<int>(arrNumber + 2, arrNumber + 6, 16);

std::cout << "i_first: " << i_first << std::endl;
std::cout << "i_second: " << i_second << std::endl;
return 0;
}
`````` ## ( Regex ) How can i convert a number to a roman form if the number is repetitive number

By : Michael
Date : March 29 2020, 07:55 AM
help you fix your problem I don't know Java very well, but I do know regular expressions, C# and JavaScript. I am confident you can adapt one of my techniques to Java.
I have sample code with two different techniques.
code :
``````private void btnRegexRep_Click(object sender, RoutedEventArgs e) {
string fixThis = @"Hans4444müller,Mary555kren";
var re = new Regex("\\d+");

string result = "";
int lastIndex = 0;
string lastMatch = "";
//Get the first match using the regular expression:
var m = re.Match(fixThis);

//Keep looping while we can match:
while (m.Success) {
//Get length of text between last match and current match:
int len = m.Index - (lastIndex + lastMatch.Length);
result += fixThis.Substring(lastIndex + lastMatch.Length, len) + GetRomanText(m);

//Save values for next iteration:
lastIndex = m.Index;
lastMatch = m.Value;
m = m.NextMatch();
}

//Append text after last match:
if (lastIndex > 0) {
result += fixThis.Substring(lastIndex + lastMatch.Length);
}

Console.WriteLine(result);
}

private string GetRomanText(Match m) {
string[] roman = new[] { "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX", "IX" };
string result = "";
// Get ASCII value of first digit from the match (remember, 48= ascii 0, 57=ascii 9):
char c = m.Value;
if (c >= 48 && c <= 57) {
int index = c - 48;
result = roman[index];
}

return result;
}
`````` ## C# How to convert text formatted number with exclamation mark icon note to number formatted number?

By : Ashish
Date : March 29 2020, 07:55 AM
I hope this helps you . Changing the number format of the cells won't do anything, since it only affects the visual presentation of a numeric value and the values are strings, not numbers.
You will need to convert the numbers from strings to a compatible numeric type - double, or int depending on the content, maybe even Decimal - and replace the string values with the converted numeric. Related Posts : 