ios,objective-c,numbers,int,digits

There are probably better solutions, but this one is slightly shorter: int theNumber = 204398234; int theDigitPlace = 3;//hundreds place int theDigit = (theNumber/(int)(pow(10, theDigitPlace - 1))) % 10; In your case, it divides the number by 100 to get 2043982 and then "extracts" the last decimal digit with the...

string,powershell,split,wildcard,digits

Use a regex with a capture group: .*?S.*?(\d{2}).*?E.* > "some.text.S**01**E02.partofstring.mkv" -replace '.*?S.*?(\d{2}).*?E.*','$1' 01 > "some.textstring.S**01**E02.partofstring.mkv" -replace '.*?S.*?(\d{2}).*?E.*','$1' 01 ...

Have you tried: bool IsCorrectIP = ( ipstring.StartsWith("10.80.") ); Sorry if the answer is too terse. But that should resolve the issue at hand....

c,constants,c-preprocessor,digits

You're missing quotes around the string. In C, a string is actually specified by a sequence of string literals. So you can do: #define DIGITS "10" printf( "%0" DIGITS "d\n", myvalue ); The preprocessor can also generate strings from other tokens (such as numbers). Check into the stringize operator (#),...

Do both at the same time There is no need for a second check. Regex allows you to check both that the folder names contains natives, and that natives is followed by a dash and 14 digits, in a single pass. Use this regex: natives-\d{14} Or, if the 14 digits...

Read the number, use an array of size 10 to count the occurrences, loop over the cyphers: int n; std::cin >> n; int occ[10] = { 0 }; while (n > 0) { const int c = n % 10; occ[c]++; n /= 10; } at the end, if occ[i]...

python,sum,digits,sum-of-digits

Use integer division instead of floating point: x //= 10 ...

An awk: awk '/^[0-9]+\.[0-9]+/{printf "\n"}{printf $0}' filename For handling DOS line breaks: awk '{sub(/\r$/,"")}/^[0-9]+\.[0-9]+/{printf "\n"}{printf $0}' filename Demo: $ awk '{sub(/\r$/,"")}/^[0-9]+\.[0-9]+/{printf "\n"}{printf $0}' filename 0 15.239 23.917 Reprenem el debat que avui els oferim entorn de les perspectives d'aquest dos mil set. <ehh> Estavem parlant concretament dels temes 30.027 de...

It seems like your toolchain has a 32-bit int type. The maximum value representable in such a type is 231-1, or 2,147,483,647. As you can see, that's a 10-digit number. You'll need to use a different type that supports larger numbers if you want to use this kind of an...

Convert to integer? static int GetDigit(int number, int k) { // k is the positiong of the digit I want to get from the number // I want to divide integer number to 10....0 (number of 0s is k) and then % 10 // to get the last digit of...

java,string,digits,significant-digits

If in your current locale the decimal separator is a dot then you will get 0.920. If you want to get the result independent from you current locale to have as decimal separator a comma and as thousand separator a dot you could achieve it for example like this String...

c++,boost,standards,bit,digits

From this std::numeric_limits::digits reference: The value of std::numeric_limits::digits is the number of digits in base-radix that can be represented by the type T without change. For integer types, this is the number of bits not counting the sign bit. And later it states that for char the result is CHAR_BIT...

What's important to realize is that it's easy to take big steps: 1 digit numbers: 123456789 - 9 * 1 digit 2 digit numbers: 101112...9899 - 90 * 2 digits 3 digit numbers: 100101102...998999 - 900 * 3 digits 4 digit numbers: ... Now you can do a recursive solution...

Your stopping condition is wrong: q%10!=0 will become "true" as soon as you reach the first zero in a decimal representation. For example, for a number 6540321 your program would add 32+22+12, and stop, because the next digit happens to be zero. Squares of 6, 5, and 4 would never...

We can write the numbers you are looking for like this: re_n = (?:[^x]|^)\d\d\d(?:[^ip]|$) Then the whole expression is: ^(?!.*re_n.*re_n.*$).*(re_n) which basically eliminates double numbers using a negative lookahead following the line start anchor, then matches a valid number. The interpolated expression looks ugly: /^(?!.*(?:(?:[^x]|^)\d\d\d(?:[^ip]|$)).*(?:(?:[^x]|^)\d\d\d(?:[^ip]|$)).*$).*((?:(?:[^x]|^)\d\d\d(?:[^ip]|$)))/ This Perl code: my $re_n...

c++,fstream,truncate,floating-point-precision,digits

The following works fine on my system (Win7, VS2012): #include <fstream> #include <iostream> int main (void) { std::ifstream file ("test.txt") ; long double d = 0 ; file >> d ; std::cout.precision (20) ; std::cout << d << "\n" ; return 0 ; } The text file: 2.7239385667867091 The output:...

sed for printing lines starting with capital letters followed by digits. It also adds a - between them: sed -n 's/^\([A-Z]\+\)\([0-9]\+\) .*/\1-\2/p' input Gives: NUC-320 CJ-101 TECH-201 ...

This should do the trick: def format(d: Double) = BigDecimal(d).scale match { case x if x > 2 => "%.3f".format(d) case _ => d.toInt.toString } ...

python,string,list,format,digits

Apply the same zfill function in a list comprehension, like this >>> [str(item).zfill(6) for item in data] ['000001', '000010', '000313', '004000', '051234', '123456'] Alternatively, you can use the string's format method, with format specifiers, like this >>> ["{:06d}".format(item) for item in data] ['000001', '000010', '000313', '004000', '051234', '123456'] If you...

You can extract the numbers: String numberStr = yourStr.replaceAll("[^0-9]", ""); if you want to keep decimal point use: yourStr.replaceAll("[^\\.0123456789]",""); then you can check for individual numbers. To add space between numbers: String numberStr = yourStr.replaceAll("[^0-9]", " "); To get only the numbers, and store them to a string array: String...

The best way I know is using forward-pipe operator %>% from magrittr package > library(magrittr) > mean(c(0.34333, 0.1728, 0.5789)) [1] 0.36501 > mean(c(0.34333, 0.1728, 0.5789)) %>% round(3) [1] 0.365 I should also mention, that package magrittr is used by another very popular package dplyr, which provides additional functionality for data...

You can use the decimal formatter for this as- NumberFormat formatter = new DecimalFormat("#0.00"); Log.d("formatted no is ",""+formatter.format(4.0)); Edit You can modify your code like this- public void onButtonClick(View v) after some formulas velocity2 = velocity * 0.00508; drop2 = pressuredrop * 249.174; vel.setText(formatter.format(velocity2)); dr.setText(formatter.format(drop2)); } ...

#include <stdio.h> int main(void) { int i, n; for (n = 1; n < 1000000; n++) { for (i = n;;) { if (i / 10 % 10 > i % 10) break; if ((i /= 10) == 0) { printf("%d\n", n); break; } } } } 5004 numbers in...

Your while condition is Math.Abs(n) > 1, but in the case of 10, you are only greater than 1 the first time. You could change this check to be >=1 and that should fix your problem. do { n = n / 10; i++; } while(Math.Abs(n) >= 1); ...