strcmp & strncmp functions return 0 if strings are equal. You should do: if (strncmp(frase1, frase3, 4) == 0) … i.e.: char *str1 = “Example 1”; char *str2 = “Example 2”; char *str3 = “Some string”; char *str4 = “Example 1”; if (strncmp(str1, str2, 7) == 0) printf(“YES\n”); // “Example” <-> “Example” else printf(“NO\n”); if … Read more
Not tested with data set but checking to the source codes of compareTo and matches it seems compareTo would be faster as matches then again calls Pattern.matches and includes other function calls as well (since matches takes regex as input – regex compilation puts extra overhead. Though since matches accepts regex as well it has … Read more
This not a integer value its a string so compare it like as Use the below function in your condition: if(!(CK.equals(uid))){ Intent intent = new Intent(LoginActivity.this, MainActivity.class); startActivity(intent); finish(); } 0 solved What is the probleme please ? [duplicate]
The amount of downvotes is crazy but oh well… I found the reason for my performance issue / bottleneck thanks to the comments. The second for loop inside StartSimilarityCheck() iterates over all entries, which in itself is no problem but when viewed under performance issues and efficient, is bad. The solution is to only check … Read more
Assuming Benoît Latinier’s interpretation of your question is right (which, it looks like it is), then there will be some cases where a unique letter can’t be found, and in these cases you might throw an exception: def unique_chars(words): taken = set() uniques = [] for word in words: for char in word: if char … Read more
patientsList = [[‘sagar’,’9856782311′],[‘mahsh’,’7865423158′]] search = input(“Enter mobile number of patient to search: “) for patient in patientsList: ”’ patient looks like this [‘sagar’,’9856782311′] patient[0] is the name patient[1] is the phone number ”’ if search == patient[1]: print(“Required patient is”) print(patient) print(“is Appointed”) break 9 solved I have given nested list accessing via for loop, … Read more