c - Allocate memory to char pointer array as it is read in -


i need read white space separated words stdin, use emalloc function allocate memory each word read in.

i finding confusing, here have written far.

#include <stdio.h> #include <stdlib.h>  int main (void) {     #define size 100     char* username[100];      int i;     int p;      /* read words array */     for(i = 0; < 100; i++)     {         username[i] = calloc(size, sizeof(char));          scanf("%s",username[i]);     }      /* print out array */     (p = 0; p < 100; p++) {         printf("%s", username[p]);     }      return 0; }

i not sure whether reading words in correctly using scanf , pretty sure memory allocation not quite correct. coming java, memory allocation tricky wrap mind around. how come not include & infront of username[i] in scanf function?

your code fine few problems:

  1. you should free'ing memory array of pointers point to.

  2. the use of scanf() dangerous buffer overflow occur.

#include <stdio.h> #include <stdlib.h> #include <string.h>  int main (void) {     #define size 100     char* username[100];      char *nl;      int i;     int p;      /* read words array */     for(i = 0; < 3; i++)     {         username[i] = calloc(size, sizeof(char));          printf("enter word %d:", i+1);         fgets( username[i], size, stdin );         // remove newline         if ((nl = strchr(username[i], '\n')))         {             *nl = '\0';         }     }      /* print out array */     (p = 0; p < 3; p++) {         printf("[%s]\n", username[p]);         free( username[p] );     }      return 0; }

output:

~/src/svn/misc > ./a.out  enter word 1:one enter word 2:two enter word 3:three 1 2 3

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