String and pointer

 C-String and Pointer

C-string (of the C language) is a character array, terminated with a null character '\0'. For example,

// Testing C-string
#include
#include
using namespace std;

int main() {
   char msg1[] = "Hello";
   char *msg2 = "Hello";
      // warning: deprecated conversion from string constant to 'char*'

   cout << strlen(msg1) << endl;    // 5
   cout << strlen(msg2) << endl;
   cout << strlen("Hello") << endl;

   int size = sizeof(msg1)/sizeof(char);
   cout << size << endl;  // 6 - including the terminating '\0'
   for (int i = 0; msg1[i] != '\0'; ++i) {
      cout << msg1[i];
   }
   cout << endl;

   for (char *p = msg1; *p != '\0'; ++p) {
          // *p != '\0' is the same as *p != 0, is the same as *p
      cout << *p;
   }
   cout << endl;
}
Take NOTE that for C-String function such as strlen() (in header cstring, ported over from C's string.h), there is no need to pass the array length into the function. This is because C-Strings are terminated by '\0'. The function can iterate thru the characters in the array until '\0'.
For example,
// Function to count the occurrence of a char in a string
#include
#include
using namespace std;

int count(const char *str, const char c);  // No need to pass the array size

int main() {
   char msg1[] = "Hello, world";
   char *msg2 = "Hello, world";

   cout << count(msg1, 'l') << endl;
   cout << count(msg2, 'l') << endl;
   cout << count("Hello, world", 'l') << endl;
}

// Count the occurrence of c in str
// No need to pass the size of char[] as C-string is terminated with '\0'
int count(const char *str, const char c) {
   int count = 0;
   while (*str) {   // same as (*str != '\0')
      if (*str == c) ++count;
      ++str;
   }
   return count;
}

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