动态 memory 分配字符串并将此字符串放置在 char 数组中

Question

I want to have an array of strings and the user to enter a string at a time. The program should either end if the the array is full or when the user skips an input (so the string would be equal to "\n". Problem is that I have to dynamically allocate memory for each of these strings and I cant find a way to do that efficiently.

Excuse my English on this one but the array should be an array of pointers to char (for example char *pin[MAX]) This is my code:

#include <stdio.h> 
#include <stdlib.h>
#include <string.h>
#define MAX 5

int main()
{
   char *pin[MAX];

   char s[] = "";
   int n = 0;

   while(s != "\n"){

       printf("Enter a string: ");
       gets(s);

       pin[n] = malloc(sizeof(char)*strlen(s));
       strcpy(pin[n], s);

       n++;
       if(n = MAX - 1) break;
   }

   for(int i = 0; i < MAX; i++){
       printf("%s ", *pin[i]);
   }
   return 0;
}

Answer 1

• Take input with fgets and store it in a temporary buffer (128 or 256 bytes large etc).
• Call strlen on the read string stored in this buffer to see how much to allocate.
• Allocate memory with malloc for pointer pin[n] and strcpy the string there.

NOTE:

• char *s; ... while(s != is nonsense since s has not been initialized.
• s != "\n" is nonsense since that's not how you compare strings in C.
• pin[n] == &s; is nonsense because it's just random stuff typed out without the programmer knowing why. Programming by trial & error doesn't work.
• In general you need to study arrays and pointers before strings.