(strlen(str) == 0) ? '\0' : str[strlen(str)-1]
String in C is basically a Character 1-D array, whose last character is a NULL ['\0']. It is declared as follows: char array_name[size]; Ex. char wiki[10];
char *ptr;
In C programming, a string doesn't have a specific return type as it's essentially an array of characters. So, if a function is returning a string, it should be declared to return a pointer to a char (char*), since a string in C is represented as an array of characters terminated by a null character ('\0').
char *string = "this is a test"; char *p; for (p=string; *p!='\0'; p++) if (*p==' ') *p='-';
Any character can be used in string, except for \\0. char example [] = "A&B|C";
Character.toString('c')
ISO C forbids converting a string constant to 'char' in C because string constants are read-only and attempting to modify them through a 'char' pointer can lead to undefined behavior and potential program crashes.
String in C is basically a Character 1-D array, whose last character is a NULL ['\0']. It is declared as follows: char array_name[size]; Ex. char wiki[10];
C forbids converting a string constant to 'char' because string constants are stored in read-only memory, and attempting to modify them through a 'char' pointer can lead to undefined behavior and potential program crashes.
Not really sure what you mean by symbol. I will assume that you are talking about a character Here is a method that will seperate each char of a String in an array of char public char[] seperateChar(String s) { char [] c = new char[s.length]; for(int i = 0; i < s.length(); i++) { c[i] = s.charAt(i); } return c }
char *ptr;
In C programming, a string doesn't have a specific return type as it's essentially an array of characters. So, if a function is returning a string, it should be declared to return a pointer to a char (char*), since a string in C is represented as an array of characters terminated by a null character ('\0').
char *string = "this is a test"; char *p; for (p=string; *p!='\0'; p++) if (*p==' ') *p='-';
Any character can be used in string, except for \\0. char example [] = "A&B|C";
Well, if you write like char a=string; it is wrong. You have to declare the size of the array or else put the brackets immediately after the variable declaration. You also have to put the string in quotes, or provide a comma-separated list of characters. E.g.,char a[]={'s','t','r','i','n','g'};Or more simply:char a[] = "string";Remember that C/C++ is case-sensitive and that all keywords are lower case. Thus Char would be regarded as an invalid keyword.
A std::string is an object that encapsulates an array of type char whereas a C-style string is a primitive array with no members. A std::string is guaranteed to be null-terminated but a C-style string is not.
//String Concatination#include#includeusing namespace std;char* strcat(char*,char*);int main(){char str1[100];char str2[100];coutstr1;coutstr2;cout