I think you meant something like “*” (star) as a wildcard for example: “a*b” => everything that starts with “a” and ends with “b” “a*” => everything that starts with “a” “*b” => everything that ends with “b” “*a*” => everything that has an “a” in it “*a*b*”=> everything that has an “a” in it, … Read more
How about using the C++ language itself? UPDATE: If you want more than 4 lines of code without any console output, please go to cppreference.com’s page talking about std::boolalpha and std::noboolalpha which shows you the console output and explains more about the API. Additionally using std::boolalpha will modify the global state of std::cout, you may want to restore the original behavior go here for … Read more
The last line was creating a problem. lastIndex would never be at -1, so there would be an infinite loop. This can be fixed by moving the last line of code into the if block.
What about: (changed to a list comp from a generator expression as using a list comp inside join is faster)
The same notation is used for pointing at a single character or the first character of a null-terminated string: The values in ptr2 and ptr3 are the same; so are the values in ptr4 and ptr5. If you’re going to treat some data as a string, it is important to make sure it is null terminated, and that you know how much … Read more
A simple example that shows how to use multiple delimiters and potential improvements in your code. See embedded comments for explanation. Be warned about the general shortcomings of strtok() (from manual): These functions modify their first argument. These functions cannot be used on constant strings. The identity of the delimiting byte is lost. The strtok() function uses a static … Read more
You can just go for String replace method.-
All floats or non string data types must be casted to strings before concatenation This should work correctly: (notice the str cast for multiplication result) straight from the interpreter:
Why this is not possible: Because String and Integer are not in the same Object hierarchy. The casting which you are trying, works only if they are in the same hierarchy, e.g. In this case, (A) objB or (Object) objB or (Object) objA will work. Hence as others have mentioned already, to convert an integer to string use: String.valueOf(integer), or Integer.toString(integer) for primitive, … Read more