Use substr. The first value is the string, the second is the starting character position (in your case, 2 is in position 0, the – is in position 1, etc.). You can also specify a length as a third parameter if you so choose. $s = substr($str, 2); solved PHP: get every character of a … Read more
It’s more than likely setting the byte to 0xFF (which is equal to -1 as a signed byte). solved In c++ what does this mean: for string n and int l n[l]=-1; [closed]
Something like this: string a = “C:/aaa/bbb/eee.txt”” ; string b = a.Replace(“C:/aaa/bbb/”, “OUT:/”); solved C# Replace variable (direction) to single word ( mapping ) [closed]
You can use regex. Here is an example: import re s=”welcome” new_s = re.sub(‘([aeiou])’, ‘\g<1>p\g<1>’, s) print(new_s) > wepelcopomepe 2 solved Python inserting characters in strings next to a specefic letters
This could be done easily using a regular expression but I am guessing you do not wish to use that. Instead the problem can be solved by reading the file in a line at a time and deciding if the line starts with a number followed by a .. If it does, start you start … Read more
The reason for the warnings and error is that strncpy() accepts char * arguments, which is different from unsigned char *. Assuming the two arrays in the structs are the same size simply do memcpy(d1.d.something, c->do_something, sizeof(d1.d.something)); If you can’t assume the two arrays are the same size, you’ll need to write code to check … Read more
Use a regular expression. import re result = re.sub(r'[dust]’, ”, string) The regexp [dust] matches any of those characters, and all the matches are replaced with an empty string. If you want to remove only the whole word dust, with possible repetitions of the letters, the regexp would be r’d+u+s+t+’. If only u can be … Read more
Introduction This article will discuss how to remove all occurrences of any characters in the word ‘dust’ from a given string. We will look at various methods of doing this, including using regular expressions, looping through the string, and using the replace() method. We will also discuss the advantages and disadvantages of each approach. Finally, … Read more
Your for loop condition is backwards. It should be i <= j. 1 solved Strings:Not able to change contents of the string in C++ [closed]
You should try this number.include?(digit.to_s) You didn’t convert the digit to string. Hope this solves your problem. 1 solved How to check whether a string includes a given string variable?
strtotime(‘2012W01 – 2 days’) for Saturday and strtotime(‘2012W01 – 1 day’) for Sunday since the week call is always going to return you a Monday. 2 solved what is the most efficient way to get 1 or 2 day before the first day of week number and year in php with native functions [closed]
You can do something like this: Pattern pat = Pattern.compile(“((?:[^\u0003]*\u0003){30}[^\u0003]*)\u0003(.*)”); Matcher matcher = pat.matcher(input); if (matcher.matches()) { String leftPart = matcher.group(1); String rightPart = matcher.group(2); } The pattern works as follows: The first capture group finds a substring with exactly 30 occurrences of the separator \u0003; the subpattern that finds a substring with no separators, … Read more
Introduction The question of how to get the day before the first day of a week number and year in PHP with native functions is a common one. Fortunately, there is an efficient way to do this using the native functions of PHP. This article will explain the most efficient way to get 1 or … Read more
Introduction String manipulation is a common task in programming, and Java provides a number of ways to split a string at a specific separator. In this article, we will discuss how to split a string at the n’th occurrence of a specific separator in Java. We will look at various examples and discuss the different … Read more
It prints “Hello” because operator << has an overload for const char* (which is what you’re passing if you pass x) that prints out a single char and moves to the next char until a NUL-character is found. “Hello” has a compiled-added NUL-character at the end, so your string is actually “Hello\0”. To get the … Read more