why use * to define string array?-CodePudding

why we need *names[] rather than names[] here, when I define it as const char names[],it will not execute.

#include <stdio.h>
#include <stdlib.h>
const int MAX = 4;
int main()
{
  const char *names[] = {
                 "dggg",
                 "ggq",
                 "gg2",
                 "g23",
   };


   for ( int i = 0; i < MAX; i  )
   {
      printf("Value of names[%d] = %s\n", i, names[i] );
   }
   return 0;
}

CodePudding user response：

A string in C is a = one dimensional character array. For instance

const char string[3] = "abc";

What the example code has is an array of C strings i.e.

{dggg", "ggq",....}

Without the pointer case, it would be like

#include <stdio.h>
#include <stdlib.h>
//const int MAX = 4;

#define NUMBER_OF_STRING 4
#define MAX_STRING_SIZE 40

int main()
{

  const char names[NUMBER_OF_STRING][MAX_STRING_SIZE] = {
                 "dggg",
                 "ggq",
                 "gg2",
                 "g23",
   };


   for ( int i = 0; i < NUMBER_OF_STRING; i  )
   {
      printf("Value of names[%d] = %s\n", i, names[i] );
   }
   return 0;
}