Gcc 7.1 -O3 these are all identical, godbolt.org/z/a-hc1d – jterm Apr 25 at 3:27 Original answer: Did some digging. Below is an example program and the relevant assembly: Code: Assembly: This was compiled with -std=c++11 -O2. You can see that there is quite a lot more work for the return “”; statement and comparably little for return std::string and return {}; (these … Read more
It’s because any iterable can be joined (e.g, list, tuple, dict, set), but its contents and the “joiner” must be strings. For example: Using something other than strings will raise the following error:
It’s because any iterable can be joined (e.g, list, tuple, dict, set), but its contents and the “joiner” must be strings. For example: Using something other than strings will raise the following error:
writelines expects an iterable of strings write expects a single string. line1 + “\n” + line2 merges those strings together into a single string before passing it to write. Note that if you have many lines, you may want to use “\n”.join(list_of_lines).
For what it’s worth, here’s another way to extract tokens from an input string, relying only on standard library facilities. It’s an example of the power and elegance behind the design of the STL. Instead of copying the extracted tokens to an output stream, one could insert them into a container, using the same generic copy algorithm. … Read more
The difference here is that will place “Hello world” in the read-only parts of the memory, and making s a pointer to that makes any writing operation on this memory illegal. While doing: puts the literal string in read-only memory and copies the string to newly allocated memory on the stack. Thus making legal.
Use to make the letter optional. {1} is redundant. (Of course you could also write [A-Z]{0,1} which would mean the same, but that’s what the ? is there for.) You could improve your regex to And, since in most regex dialects, \d is the same as [0-9]: But: do you really need 11 separate capturing groups? And if so, why don’t you capture the … Read more
Use str.join:
“N/A” is not an integer. It must throw NumberFormatException if you try to parse it to an integer. Check before parsing or handle Exception properly. Exception Handlingtry{ int i = Integer.parseInt(input); } catch(NumberFormatException ex){ // handle your exception … } or – Integer pattern matching –
You can use the localeCompare() method.