Your’re probably not using it in a method. Place it in a method as shown below : public static void main(String[] args) { char aCharacter=”A”; aCharacter=”\u0041″; } This is because assignments are statements and statements are allowed only inside blocks of code. 0 solved When assigning a value Java gives the error ‘java: expected’, Why? … Read more
If you want to find the maximum sub array length which contains only continious r and b, here is a solution. The basic idea is using two cursor and greedy search. public static int findMaximum(char[] input) { int result = 0; int first = 0; int second = 0; while (input[first] == input[second]) { second++; … Read more
The expression int(ean8[-1]) takes the last character [-1] of the string ean8 and converts it into an integer to allow further operations, that most require an integer (instead of a string) to be executed. This is due to the fact that the barcode is present in form of a string, or a sequence of characters. … Read more
a is of type char and chars can be implicitly converted to int. a is represented by 97 as this is the codepoint of small latin letter a. System.out.println(‘a’); // this will print out “a” // If we cast it explicitly: System.out.println((int)’a’); // this will print out “97” // Here the cast is implicit: System.out.println(‘a’ … Read more
There are many uses of ! operator in Linux. See this article. In your case, it is probably to check if the a command from history is given as input. See the first point of the article. solved ‘!’ in input buffer in comparison in if statement
Done! Credit : @AlgirdasPreidžius #include <iostream> using namespace std; int fun() { string c; while(cin>>c) { if(isdigit(c[0])) return stoi(c); } } int main() { int a = fun(); cout<<a<<” “; return 0; } 3 solved Getting input from user till he enters a number
The difference is: names1 is declared simply as an array of string pointers without other defined characteristics. Using string literals here will put the string literals into a section in your executable file which is read-only, because the compiler can this way re-use them. For example, when you use char* a = “abc”; char* b … Read more
const means constant, so const char* means a constant char pointer, or a char pointer that cannot be modified, but a char can be modified because it is not a constant, but can only hold one character because it isn’t a pointer (a.k.a. a char array). Anyway, have you tried combining the two variables with … Read more
Try the Scanner class. Like this: import java.util.Scanner; public class New { public static void main(String[] args) { System.out.println(“Calculator”); Scanner scanner = new Scanner(System.in); System.out.println(“Enter Parameter “); System.out.print(“a : “); float a = scanner.nextFloat(); System.out.print(“+|-|*|/: “); String op = scanner.next(); System.out.print(“b : “); float b = scanner.nextFloat(); float c = 0; switch (op) { case … Read more
It looks like you are trying to build a string. For that you can use snprintf. char sql[64]; int size = snprintf(sql, 64, “SELECT %lf/(%lf * %lf);”, peso, altura, altura); Note that I kept the return value. This is to handle any future error where you modify a statement in a way that might result … Read more
The first parameter to fputs is a char*, so the code you show is obviously incorrect. You say I tried itoa or sstream functions but it’s not working. but those are the solutions, and there’s no reason for them not to work. int a = 5; //the C way FILE* pFile = fopen(“myfile.txt”,”w”); char buffer[12]; … Read more
You can delete only what was allocated using the corresponding operator new. String literals have static storage duration. They are not dynamically allocated. According to the C++ Standard 8 Ordinary string literals and UTF-8 string literals are also referred to as narrow string literals. A narrow string literal has type “array of n const char”, … Read more
In C and C++, regular character strings are terminated by a ‘\0’ character. So it appears you are using a string function to copy the data. That is, it stops copying once it hits the ‘\0’ value. Interestingly, you left the part that copies the data (and maybe just the part the displays the data) … Read more
You are confusing strings and characters. A character literal is enclosed in ‘. A string literal is enclosed in ” and is an array of characters. e.g.: “This is a string literal” ‘c’ // this is character To test if a char is not z for instance, you write: char c = …; if (c … Read more
These both statements are wrong char[] name = { “Nitish prajapati” }; char* namePointer = &name ; In C++ valid declaration of an array looks like char name[] = { “Nitish prajapati” }; As for the second statement then there is no implicit conversion from type char ( * )[17] to char *. The initializer … Read more