See CppReference operator [] No bounds checking is performed. If pos > size(), the behavior is undefined. When you define string r, its size() is zero, so all your character assignments are undefined behaviors. You may want to resize it first: r.resize(9); Or alternatively, append the characters one-by-one: r = “”; r.push_back(hh/10+’0′); r.push_back(hh%10+’0′); r.push_back(‘:’); r.push_back(mm/10+’0′); … Read more
You are trying to change List<String> to List<tcmb> and problem lies here public static List<tcmb> getList(Activity a){ // your code // List<String> itemList = new ArrayList<String>(); itemList.addAll(Arrays.asList(itemWords)); dovizList = (List)itemsList; Log.d(TAG, “getValuestcmb: ” + dovizList.size()); return dovizList; Also, I don’t understand, what exactly you are trying to achieve here. List<String> itemsList = new ArrayList<String>(); dovizList … Read more
Given a list, you can use ‘ ‘.join: s = [‘hello’,’world’] new_s=” “.join(s) Output: ‘hello world’ 1 solved How to change a split str into a normal str [duplicate]
A better way to achieve what you’re trying to do overall would be to use an exception. Consider if we wrote AddNumbers as such: public static int AddNumbers(int number1, int number2) { int result = number1 + number2; if(result <= 10) { throw new Exception(“Should be more than 10”); } return result; } and our … Read more
Based on @Kevin s suggestion with “ast.parse“ I could use def hasRecursion(tree): for node in [n for n in ast.walk(tree)]: nodecls = node.__class__ nodename = nodecls.__name__ if isinstance(node, (ast.For, ast.While)): for nodeChild in node.body: #node.body Or ast.iter_child_nodes(node) if isinstance(nodeChild, (ast.For, ast.While)): return True return False expr=””” for i in 3: print(“first loop”) for j in … Read more
Use the .at() function of the string string a = “ATest”; if(a.at(0) == ‘A’) cout<<“True”; else cout<<“False”; 5 solved How Do I Test If The First Letter Of A String Is “A” in C++?
I must say thanks to user853710 for a quick pattern. This will help you to collect the date from the given sting, and DateTime.TryParseExact plays the key role in the extraction process: string pattern = @”\d{4}-\d{2}-\d{2} \d{1,2}:\d{1,2}:\d{1,2}”; string p = “v 0 fl 0x4; value 8 feat 0; sn AC0809099; mn -; tim 2015-10-11 20:50:36 … Read more
split your string on space to get Logging and v0.12.4 remove (substring) v from v0.12.4 split 0.12.4 on dot (use split(“\\.”) since dot is special in regex) you can also parse each “0” “12” “4” to integer (Integer.parseInt can be helpful). 4 solved Parsing String and Int in java
Your question seems to be more about PHP than Javascript, based on the code snippet you provided. Basically, you might want to escape the apostrophes with a backslash character. onclick=”(method(‘<?php echo $variable; ?>’))” would then become something like this perhaps: onclick=”(method(‘<?php echo addslashes($variable); ?>’))” … and please remove the parenthesis you have surrounding the onclick … Read more
Your problem lies in the use of consecutive if statements without chaining them together to make a complete check. Doing it your way, the code actually completely disregards all the if statements, but the last one. So, if iledefrance.indexOf(departement) != -1 gives false, it’will always execute the code inside else, meaning it’ll set region = … Read more
Writing a function to print letters of a string according to their frequency compalsarily using dictionary and tuple [closed] solved Writing a function to print letters of a string according to their frequency compalsarily using dictionary and tuple [closed]
you should import your Module or external library first. and random module dose not have random.str() function! Try This : import random a = [‘blue’, ‘red’, ‘green’, ‘yellow’, ‘brown’, ‘black’] print(random.choice(a)) Good Luck. 1 solved How to generate a random word
You should certainly read more on regex. A first hint is that when you want to capture a pattern you need to enclose it in parentheses. e.g. (\d+). For this example though, the code you need is: match = re.match(r’C(\d+)([F|G])(\d+)\.PNG’, s) first_id = match.group(1) fg_class = match.group(2) second_id = match.group(3) 1 solved How to extract … Read more
ex.message.Contains(“Cannot open server”) && ex.message.Contains(“requested by the login”) solved How can I check if a string contains two other strings [closed]
%8.1f is a format specifier. This should point you in the right direction for figuring out what all the parts mean. It specifies the format in which a variable will be printed. In this specific case, it is a place-holder for a floating point variable. It reserves a width of at least 8 characters and … Read more